NCERT Solutions for Class 11 Maths Chapter 10 – Straight Lines

ANSWER:

Let ABC be the given equilateral triangle with side 2a.

Accordingly, AB = BC = CA = 2a

Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.

i.e., BO = OC = a, where O is the origin.

Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex A lies on the y-axis.

On applying Pythagoras theorem to ΔAOC, we obtain

(AC)² = (OA)² + (OC)²

⇒ (2a)² = (OA)² + a²

⇒ 4a² – a² = (OA)²

⇒ (OA)² = 3a²

⇒ OA =

∴Coordinates of point A =

Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and  or (0, a), (0, –a), and.

ANSWER:

The given points are and.

(i) When PQ is parallel to the y-axis, x₁ = x₂.

In this case, distance between P and Q 

(ii) When PQ is parallel to the x-axis, y₁ = y₂.

In this case, distance between P and Q 

ANSWER:

Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

On squaring both sides, we obtain

a² – 14a + 85 = a² – 6a + 25

⇒ –14a + 6a = 25 – 85

⇒ –8a = –60

Thus, the required point on the x-axis is.

ANSWER:

The coordinates of the mid-point of the line segment joining the points

P (0, –4) and B (8, 0) are

It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by.

Therefore, the slope of the line passing through (0, 0) and (4, –2) is

.

Hence, the required slope of the line is.

ANSWER:

The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).

It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by.

∴Slope of AB (m₁)

Slope of BC (m₂)

Slope of CA (m₃)

It is observed that m₁m₃ = –1

This shows that line segments AB and CA are perpendicular to each other

i.e., the given triangle is right-angled at A (4, 4).

Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

ANSWER:

If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° 

ANSWER:

If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then

Slope of AB = Slope of BC

Thus, the required value of x is 1.

ANSWER:

Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

Slope of AB

Slope of CD =

⇒ Slope of AB = Slope of CD

⇒ AB and CD are parallel to each other.

Now, slope of BC =

Slope of AD =

⇒ Slope of BC = Slope of AD

⇒ BC and AD are parallel to each other.

Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.

Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

ANSWER:

The slope of the line joining the points (3, –1) and (4, –2) is 

Now, the inclination (θ ) of the line joining the points (3, –1) and (4, – 2) is given by

tan θ= –1

⇒ θ = (90° + 45°) = 135°

Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

ANSWER:

Let be the slopes of the two given lines such that.

We know that if θisthe angle between the lines l₁ and l₂ with slopes m₁ and m₂, then.

It is given that the tangent of the angle between the two lines is.

Case I

If m = –1, then the slopes of the lines are –1 and –2.

If m =, then the slopes of the lines areand –1.

Case II

If m = 1, then the slopes of the lines are 1 and 2.

If m =, then the slopes of the lines are.

Hence, the slopes of the lines are –1 and –2 or and –1 or 1 and 2 or.

ANSWER:

The slope of the line passing through is.

It is given that the slope of the line is m.

Hence,

ANSWER:

If the points A (h, 0), B (a, b), and C (0, k) lie on a line, then

Slope of AB = Slope of BC

On dividing both sides by kh, we obtain

Hence,

ANSWER:

Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is 

Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y).

∴Slope of AB = Slope of BC

Thus, the slope of line AB is, while in the year 2010, the population will be 104.5 crores.

ANSWER:

We know that the equation of the line passing through point, whose slope is m, is.

Thus, the equation of the line passing through point (–4, 3), whose slope is, is

ANSWER:

We know that the equation of the line passing through point, whose slope is m, is.

Thus, the equation of the line passing through point (0, 0), whose slope is m,is

(y – 0) = m(x – 0)

i.e., y = mx

ANSWER:

The slope of the line that inclines with the x-axis at an angle of 75° is

m = tan 75°

We know that the equation of the line passing through point, whose slope is m, is.

Thus, if a line passes thoughand inclines with the x-axis at an angle of 75°, then the equation of the line is given as

ANSWER:

It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as

y = m(x – d)

For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.

The slope of the line is given as m = –2

Thus, the required equation of the given line is

y = –2 [x – (–3)]

y = –2x – 6

i.e., 2x + y + 6 = 0

ANSWER:

It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as

y = mx + c

Here, c = 2 and m = tan 30°.

Thus, the required equation of the given line is

ANSWER:

It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is.

Therefore, the equation of the line passing through the points (–1, 1) and

(2, –4) is

ANSWER:

If p is the length of the normal from the origin to a line and ω is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by xcos ω + y sin ω = p.

Here, p = 5 units and ω = 30°

Thus, the required equation of the given line is

x cos 30° + y sin 30° = 5

ANSWER:

It is given that the vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5).

Let RL be the median through vertex R.

Accordingly, L is the mid-point of PQ.

By mid-point formula, the coordinates of point L are given by

It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is.

Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2).

Hence, 

Thus, the required equation of the median through vertex R is.

ANSWER:

The slope of the line joining the points (2, 5) and (–3, 6) is

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6) 

Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

ANSWER:

According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by

The slope of the line joining the points (1, 0) and (2, 3) is

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) 

Now, the equation of the line passing through and whose slope isis given by

ANSWER:

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

It is given that the line cuts off equal intercepts on both the axes. This means that a = b.

Accordingly, equation (i) reduces to

Since the given line passes through point (2, 3), equation (ii) reduces to

2 + 3 = a ⇒ a = 5

On substituting the value of a in equation (ii), we obtain

x + y = 5, which is the required equation of the line

ANSWER:

The equation of a line in the intercept form is

Here, a and b are the intercepts on x and y axes respectively.

It is given thata + b = 9 ⇒ b = 9 – a … (ii)

From equations (i) and (ii), we obtain

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to

If a = 6 and b = 9 – 6 = 3, then the equation of the line is

If a = 3 and b = 9 – 3 = 6, then the equation of the line is

ANSWER:

The slope of the line making an anglewith the positive x-axis is 

Now, the equation of the line passing through point (0, 2) and having a slope  is.

The slope of line parallel to line is.

It is given that the line parallel to linecrosses the y-axis 2 units below the origin i.e., it passes through point (0, –2).

Hence, the equation of the line passing through point (0, –2) and having a slope is

ANSWER:

The slope of the line joining the origin (0, 0) and point (–2, 9) is 

Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is

Now, the equation of the line passing through point (–2, 9) and having a slope m₂ is

ANSWER:

It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value of L is 125.134.

Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.

Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and (110, 125.134) in the XY plane.

Therefore, the linear relation between L and C is the equation of the line passing through points (20, 124.942) and (110, 125.134).

(L – 124.942) = 

ANSWER:

The relationship between selling price and demand is linear.

Assuming selling price per litre along the x-axis and demand along the y-axis, we have two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship between selling price and demand.

Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points (14, 980) and (16, 1220).

When x = Rs 17/litre,

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

ANSWER:

Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.

Since P (a, b) is the mid-point of AB,

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).

The equation of the line passing through points (0, 2b) and (2a, 0) is

On dividing both sides by ab, we obtain

Thus,the equation of the line is.

ANSWER:

Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.

Let the respective coordinates of A and B be (x, 0) and (0, y).

Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

Therefore, the respective coordinates of A and B are and (0, 3k).

Now, the equation of line AB passing through points  and

(0, 3k) is

Thus,the required equation of the line is 2kx + hy = 3hk.

ANSWER:

In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).

The equation of the line passing through points (3, 0) and (–2, –2) is

It is observed that at x = 8 and y = 2,

L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.

Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.

ANSWER:

(i) The given equation is 3x + 2y – 12 = 0.

It can be written as

This equation is of the form, where a = 4 and b = 6.

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6.

It can be written as

This equation is of the form, where a =  and b = –2.

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are  and –2 respectively.

(iii) The given equation is 3y + 2 = 0.

It can be written as

This equation is of the form, where a = 0 and b = .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is and it has no intercept on the x-axis.

ANSWER:

(i) The given equation is.

It can be reduced as:

On dividing both sides by, we obtain

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

x cos ω + y sin ω = p, we obtain ω = 120° and p = 4.

Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120°.

(ii) The given equation is y – 2 = 0.

It can be reduced as 0.x + 1.y = 2

On dividing both sides by, we obtain 0.x + 1.y = 2

⇒ x cos 90° + y sin 90° = 2 … (1)

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

x cos ω + y sin ω = p, we obtain ω = 90° and p = 2.

Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive x-axis is 90°.

(iii) The given equation is x – y = 4.

It can be reduced as 1.x + (–1) y = 4

On dividing both sides by, we obtain

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of line

x cos ω + y sin ω = p, we obtain ω = 315° and .

Thus, the perpendicular distance of the line from the origin is, while the angle between the perpendicular and the positive x-axis is 315°.

ANSWER:

The given equation of the line is 12(x + 6) = 5(y – 2).

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 12, B = –5, and C = 82.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

The given point is (x₁, y₁) = (–1, 1).

Therefore, the distance of point (–1, 1) from the given line

ANSWER:

The given equation of line is

On comparing equation (1) with general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

Therefore,

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

ANSWER:

It is known that the distance (d) between parallel lines Ax + By + C₁ = 0 and Ax + By + C₂ = 0 is given by.

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0.

Here, A = 15, B = 8, C₁ = –34, and C₂ = 31.

Therefore, the distance between the parallel lines is

(ii) The given parallel lines are l (x + y) + p = 0 and l (x + y) – r = 0.

lx + ly + p = 0 and lx + ly – r = 0

Here, A = l, B = l, C₁ = p, and C₂ = –r.

Therefore, the distance between the parallel lines is

ANSWER:

The equation of the given line is

, which is of the form y = mx + c

∴ Slope of the given line 

It is known that parallel lines have the same slope.

∴ Slope of the other line = 

Now, the equation of the line that has a slope of  and passes through the point (–2, 3) is

ANSWER:

The given equation of line is.

, which is of the form y = mx + c

∴Slope of the given line

The slope of the line perpendicular to the line having a slope of  is 

The equation of the line with slope –7 and x-intercept 3 is given by

y = m (x – d)

⇒ y = –7 (x – 3)

⇒ y = –7x + 21

⇒ 7x + y = 21

ANSWER:

The given lines are.

The slope of line (1) is, while the slope of line (2) is.

The acute angle i.e., θ between the two lines is given by

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

ANSWER:

The slope of the line passing through points (h, 3) and (4, 1) is

The slope of line 7x – 9y – 19 = 0 or  is.

It is given that the two lines are perpendicular.

Thus, the value of h is.

ANSWER:

The slope of line Ax + By + C = 0 or  is 

It is known that parallel lines have the same slope.

∴ Slope of the other line = 

The equation of the line passing through point (x₁, y₁) and having a slope is

Hence, the line through point (x₁, y₁) and parallel to line Ax + By + C = 0 is

A (x –x₁) + B (y – y₁) = 0

ANSWER:

It is given that the slope of the first line, m₁ = 2.

Let the slope of the other line be m₂.

The angle between the two lines is 60°.

The equation of the line passing through point (2, 3) and having a slope of is

In this case, the equation of the other line is.

The equation of the line passing through point (2, 3) and having a slope of is

In this case, the equation of the other line is.

Thus, the required equation of the other line is or .

ANSWER:

The right bisector of a line segment bisects the line segment at 90°.

The end-points of the line segment are given as A (3, 4) and B (–1, 2).

Accordingly, mid-point of AB 

Slope of AB

∴Slope of the line perpendicular to AB =

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = –2 (x – 1)

y – 3 = –2x + 2

2x + y = 5

Thus, the required equation of the line is 2x + y = 5.

ANSWER:

Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

Slope of the line joining (–1, 3) and (a, b), m₁

Slope of the line 3x – 4y – 16 = 0 or 

Since these two lines are perpendicular, m₁m₂ = –1

Point (a, b) lies on line 3x – 4y = 16.

∴3a – 4b = 16 … (2)

On solving equations (1) and (2), we obtain

Thus, the required coordinates of the foot of the perpendicular are.

ANSWER:

The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (–1, 2).

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

∴Slope of the line joining (0, 0) and (–1, 2) 

The slope of the given line is m.

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx + c.

Thus, the respective values of m and c are.

ANSWER:

The equations of given lines are

x cos θ – y sinθ = k cos 2θ … (1)

x secθ + y cosec θ= k … (2)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

On comparing equation (1) to the general equation of line i.e., Ax + By + C = 0, we obtain A = cosθ, B = –sinθ, and C = –k cos 2θ.

It is given that p is the length of the perpendicular from (0, 0) to line (1).

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain A = secθ, B = cosecθ, and C = ­–k.

It is given that q is the length of the perpendicular from (0, 0) to line (2).

From (3) and (4), we have

Hence, we proved that p² + 4q² = k².

ANSWER:

Let AD be the altitude of triangle ABC from vertex A.

Accordingly, AD⊥BC

The equation of the line passing through point (2, 3) and having a slope of 1 is

(y – 3) = 1(x – 2)

⇒ x – y + 1 = 0

⇒ y – x = 1

Therefore, equation of the altitude from vertex A = y – x = 1.

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C = –3.

∴Length of AD 

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and units respectively.

ANSWER:

It is known that the equation of a line whose intercepts on the axes are a and b is

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = –ab.

Therefore, if p is the length of the perpendicular from point (x₁, y₁) = (0, 0) to line (1), we obtain

On squaring both sides, we obtain

Hence, we showed that.

ANSWER:

The equation of the given line is.

This equation can be reduced as

On dividing both sides by, we obtain

On comparing equation (1) to, we obtain

Since the values of sin θ and cos θ are negative, 

Thus, the respective values of θand p are  and 1

ANSWER:

Let the intercepts cut by the given lines on the axes be a and b.

It is given that

a + b = 1 … (1)

ab = –6 … (2)

On solving equations (1) and (2), we obtain

a = 3 and b = –2 or a = –2 and b = 3

It is known that the equation of the line whose intercepts on the axes are a and b is

Case I: a = 3 and b = –2

In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6.

Case II: a = –2 and b = 3

In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6.

Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

ANSWER:

Let (0, b) be the point on the y-axis whose distance from line  is 4 units.

The given line can be written as 4x + 3y – 12 = 0 … (1)

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 4, B = 3, and C = –12.

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

Therefore, if (0, b) is the point on the y-axis whose distance from line  is 4 units, then:

Thus, the required points are and.

ANSWER:

The equation of the line joining the points is given by

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by.

Therefore, the perpendicular distance (d) of the given line from point (x₁, y₁) = (0, 0) is

ANSWER:

The equation of any line parallel to the y-axis is of the form

x = a … (1)

The two given lines are

x – 7y + 5 = 0 … (2)

3x + y = 0 … (3)

On solving equations (2) and (3), we obtain.

Therefore, is the point of intersection of lines (2) and (3).

Since line x = a passes through point, .

Thus, the required equation of the line is.

ANSWER:

The equation of the given line is.

This equation can also be written as 3x + 2y – 12 = 0

, which is of the form y = mx + c

∴Slope of the given line 

∴Slope of line perpendicular to the given line

Let the given line intersect the y-axis at (0, y).

On substituting x with 0 in the equation of the given line, we obtain

∴The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of and passes through point (0, 6) is

Thus, the required equation of the line is.

ANSWER:

The equations of the given lines are

y – x = 0 … (1)

x + y = 0 … (2)

x – k = 0 … (3)

The point of intersection of lines (1) and (2) is given by

x = 0 and y = 0

The point of intersection of lines (2) and (3) is given by

x = k and y = –k

The point of intersection of lines (3) and (1) is given by

x = k and y = k

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k).

We know that the area of a triangle whose vertices are (x₁, y₁), (x₂, y₂), and (x₃, y₃) is.

Therefore, area of the triangle formed by the three given lines

ANSWER:

The equations of the given lines are

3x + y – 2 = 0 … (1)

px + 2y – 3 = 0 … (2)

2x – y – 3 = 0 … (3)

On solving equations (1) and (3), we obtain

x = 1 and y = –1

Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).

p (1) + 2 (–1) – 3 = 0

p – 2 – 3 = 0

p = 5

Thus, the required value of p is 5.

ANSWER:

The equations of the given lines are

y = m₁x + c₁ … (1)

y = m₂x + c₂ … (2)

y = m₃x + c₃ … (3)

On subtracting equation (1) from (2), we obtain

On substituting this value of x in (1), we obtain

 is the point of intersection of lines (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

Hence,

ANSWER:

Let the slope of the required line be m₁.

The given line can be written as , which is of the form y = mx + c

∴Slope of the given line = 

It is given that the angle between the required line and line x – 2y = 3 is 45°.

We know that if θisthe acute angle between lines l₁ and l₂ with slopes m₁ and m₂ respectively, then.

Case I: m₁ = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y – 2 = 3 (x – 3)

y – 2 = 3x – 9

3x – y = 7

Case II: m₁ = 

The equation of the line passing through (3, 2) and having a slope of is:

Thus, the equations of the lines are 3x – y = 7 and x + 3y = 9.

ANSWER:

Let the equation of the line having equal intercepts on the axes be

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain.

 is the point of intersection of the two given lines.

Since equation (1) passes through point,

∴ Equation (1) becomes

Thus, the required equation of the line is.

ANSWER:

Let the equation of the line passing through the origin be y = m₁x.

If this line makes an angle of θ with line y = mx + c, then angle θ is given by

Case I: 

Case II: 

Therefore, the required line is given by.

ANSWER:

The equation of the line joining the points (–1, 1) and (5, 7) is given by

The equation of the given line is

x + y – 4 = 0 … (2)

The point of intersection of lines (1) and (2) is given by

x = 1 and y = 3

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. Accordingly, by section formula,

Thus, the line joining the points (–1, 1) and (5, 7) is divided by line

x + y = 4 in the ratio 1:2.

ANSWER:

The given lines are

2x – y = 0 … (1)

4x + 7y + 5 = 0 … (2)

A (1, 2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain.

∴Coordinates of point B are.

By using distance formula, the distance between points A and B can be obtained as

Thus, the required distance is.

ANSWER:

Let y = mx + c be the line through point (–1, 2).

Accordingly, 2 = m (–1) + c.

⇒ 2 = –m + c

⇒ c = m + 2

∴ y = mx + m + 2 … (1)

The given line is

x + y = 4 … (2)

On solving equations (1) and (2), we obtain

 is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

ANSWER:

Let A(1,3) and B(−4,1) be the coordinates of the end points of the hypotenuse.

Now, plotting the line segment joining the points A(1,3) and B(−4,1) on the coordinate plane, we will get two right triangles with AB as the hypotenuse. Now from the diagram, it is clear that the point of intersection of the other two legs of the right triangle having AB as the hypotenuse can be either P or Q.

CASE 1: When ∆ APB is taken.
The perpendicular sides in ∆ APB are AP and PB.
Now, side PB  is parallel to x-axis and at a distance of 1 units above x-axis.
So, equation of PB is, y=1 or y−1=0.
The side AP is parallel to y-axis and at a distance of 1 units on the right of y-axis.
So, equation of AP is x=1 or x−1=0.

CASE 2: When ∆ AQB is taken.
The perpendicular sides in ∆ AQB are AQ and QB.
Now, side AQ  is parallel to x-axis and at a distance of 3 units above x-axis.
So, equation of AQ is, y=3 or y−3=0.
The side QB is parallel to y-axis and at a distance of 4 units on the left of y-axis.
So, equation of QB is x=−4 or x+4=0.

Hence, the equation of the legs are :
x=1, y=1 or x=−4, y=3

ANSWER:

The equation of the given line is

x + 3y = 7 … (1)

Let point B (a, b) be the image of point A (3, 8).

Accordingly, line (1) is the perpendicular bisector of AB.

Since line (1) is perpendicular to AB,

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

On solving equations (2) and (3), we obtain a = –1 and b = –4.

Thus, the image of the given point with respect to the given line is (–1, –4).

ANSWER:

The equations of the given lines are

y = 3x + 1 … (1)

2y = x + 3 … (2)

y = mx + 4 … (3)

Slope of line (1), m₁ = 3

Slope of line (2), 

Slope of line (3), m₃ = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).

Thus, the required value of m is.

ANSWER:

The equations of the given lines are

x + y – 5 = 0 … (1)

3x – 2y + 7 = 0 … (2)

The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by

It is given that.

, which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of.

Thus, point P must move on a line.

ANSWER:

The equations of the given lines are

9x + 6y – 7 = 0 … (1)

3x + 2y + 6 = 0 … (2)

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by

The perpendicular distance of P (h, k) from line (2) is given by

Since P (h, k) is equidistant from lines (1) and (2), 

∴

9h + 6k – 7 = – 9h – 6k – 18

⇒ 18h + 12k + 11 = 0

Thus, the required equation of the line is 18x + 12y + 11 = 0.

ANSWER:

Let the coordinates of point A be (a, 0).

Draw a line (AL) perpendicular to the x-axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

∠BAL = ∠CAL = Φ

Let ∠CAX = θ

∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]

= 180° – θ – 180° + 2θ

= θ

∴∠BAX = 180° – θ

From equations (1) and (2), we obtain

Thus, the coordinates of point A are.

ANSWER:

The equation of the given line is

Length of the perpendicular from point to line (1) is 

Length of the perpendicular from point  to line (2) is

On multiplying equations (2) and (3), we obtain

Hence, proved.

ANSWER:

The equations of the given lines are

2x – 3y + 4 = 0 … (1)

3x + 4y – 5 = 0 … (2)

6x – 7y + 8 = 0 … (3)

The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain.

Thus, the person is standing at point.

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point.

∴Slope of the line perpendicular to line (3) 

The equation of the line passing through  and having a slope of is given by

Hence, the path that the person should follow is.