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The **“Straight Lines”** chapter of Maths Class 11 teaches general equation
of a line, slope of a line, the distance of a point from a line, various forms of
equations of a line: slope-intercept form, normal form, parallel to axes, intercepts
form, point-slope form, two-point form; the angle between two lines, and much more.

##### Question 1:

Find the slope of the line :

Passing through the points (3, –2) and (–1

##### Answer:

##### Question 2:

Find the slope of the line :

Passing through the points (3, –2) and (7, –2)

##### Answer:

##### Question 3:

Find the slope of the line :

Passing through the points (3, –2) and (3, 4)

##### Answer:

##### Question 4:

Find the slope of the line :

Making inclination of 60$\xb0$ with the positive direction
of x-axis.

##### Answer:

##### Question 5:

##### Answer:

##### Question 6:

Line through the points (–2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of ‘x’.

##### Answer:

##### Question 7:

##### Answer:

##### Question 8:

In the figure, time-distance graph of a
linear relation is given :

Two positions of time and distance recorded as :

When T = 0, D = 2 and when T = 3, D = 8.

Using the concept of slope, find the law of motion
i.e. how distance depends upon time.

##### Answer:

$\Rightarrow $ 2 (T – 3) = D – 8 $\Rightarrow $ D = 2 (T + 1), which is the reqd. relation.

##### Question 9:

Find the equations of the lines parallel to axes and passing through (–2, 3).

##### Answer:

(i) Any line parallel to x-axis is y = b. ...(1)
Since it passes through (–2, 3),

$\therefore $ 3 = b

$\Rightarrow $ b = 3.

Putting in (1), y = 3, which is the required equation.

(ii) Any line parallel to y-axis is x = a ...(1)

Since it passes through (–2, 3),

$\therefore $ –2 = a

$\Rightarrow $ a = –2.

Putting in (1), x = –2, which is the required equation.

##### Question 10:

Find the equation of line through (–2, 3) with slope – 4.

##### Answer:

The equation of the line is y – y_{1} = m(x – x_{1})

$\Rightarrow $ y – 3 = – 4[x – (–2)]

$\Rightarrow $ y – 3 = – 4(x + 2)

$\Rightarrow $ y – 3 = – 4x – 8

$\Rightarrow $ 4x + y + 5 = 0.

##### Question 11:

Write the equation of the line through the points (1, –1) and (3, 5).

##### Answer:

##### Question 12:

##### Answer:

##### Question 13:

x-intercept is 4

##### Answer:

##### Question 14:

Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively.

##### Answer:

##### Question 15:

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15$\xb0$.

##### Answer:

##### Question 16:

The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and the value of F when K = 0.

##### Answer:

##### Question 17:

Equation of a line is 3x – 4y + 10 = 0. Find its :

(i) slope (ii) x- and y-intercepts

##### Answer:

##### Question 18:

##### Answer:

##### Question 19:

##### Answer:

##### Question 20:

##### Answer:

##### Question 21:

##### Answer:

##### Question 22:

Find the equation of a line perpendicular to the line x – 2y + 3 = 0 and passing through the point (1, –2).

##### Answer:

##### Question 23:

Find the distance of the point (3, –5) from the line 3x – 4y – 26 = 0.

##### Answer:

##### Question 24:

Find the distance between the parallel
lines :

3x – 4y + 5 = 0 and 3x – 4y + 7 = 0.

##### Answer:

##### Question 25:

If the lines :

2x + y – 3 = 0, 5x + ky – 3 = 0 and 3x – y – 2 = 0
are concurrent, find the value of ‘k’.

##### Answer:

##### Question 26:

Find the distance of the line 4x – y = 0 from the point P(4, 1) measured along the line making an angle of 135$\xb0$ with the positive x-axis.

##### Answer:

##### Question 27:

Assuming that straight line works as the plane mirror for a point, find the image of the point (1, 2) in the line x – 3y + 4 = 0.

##### Answer:

##### Question 28:

##### Answer:

##### Question 29:

A line is such that its segment between the
lines :

5x – y + 4 = 0 and 3x + 4y – 4 = 0
is bisected at the point (1, 5). Obtain its equation.

##### Answer:

The given lines are ;

5x – y + 4 = 0 ...(1)

and 3x + 4y – 4 = 0 ...(2)

##### Question 30:

Show that the path of a moving point such
that its distance from the lines :

3x – 2y = 5 and 3x + 2y = 5
are equal, is a straight line.

##### Answer:

##### Question 31:

Draw a quadrilateral in the cartesian plane whose vertices are (– 4, 5), (0, 7), (5, –5) and (– 4, –2). Also find its area.

##### Answer:

Let A (– 4, 5), B (0, 7), C (5, –5) and D (–4, –2) be the vertices of the quad. ABCD, which is as shown in the figure.

##### Question 32:

The base of an equilateral triangle with side 2a lies along the y-axis such that the mid–point of the base is the origin. Find the vertices of the triangle.

##### Answer:

##### Question 33:

PQ is parallel to y-axis

##### Answer:

##### Question 34:

PQ is parallel to x-axis

##### Answer:

##### Question 35:

Find a point on the x-axis, which is equidistant from (7, 6) and (3, 4).

##### Answer:

##### Question 36:

Find the slope of the line, which passes through the origin and the mid-point of the line segment joining the points P (0, –4) and Q (8, 0).

##### Answer:

##### Question 37:

Without using Pythagorean Theorem, show that (4, 4), (3, 5) and (–1, –1) are vertices of a right triangle.

##### Answer:

##### Question 38:

Find the slope of the line which makes an angle of 30$\xb0$ with the positive direction of y-axis, measured anticlockwise.

##### Answer:

##### Question 39:

Find the value of x for which the points (x, –1), (2, 1) and (4, 5) are collinear.

##### Answer:

##### Question 40:

Without using the distance formula, show that (–2, –1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

##### Answer:

Let A (–2, –1), B (4, 0), C (3, 3) and D (–3, 2) be the given points.

$\therefore $ points A, B and C are not collinear.

Similarly, the points A, D and C are not collinear.

Thus the given points form a quadrilateral.

Now slope of AB = slope of CD

$\Rightarrow $ AB || CD

and slope of BC = slope of DA $\Rightarrow $ BC || DA.

Hence the given points form a parallelogram.

##### Question 41:

Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).

##### Answer:

##### Question 42:

##### Answer:

##### Question 43:

##### Answer:

##### Question 44:

##### Answer:

##### Question 45:

##### Answer:

##### Question 46:

find the equation of the line, which satisfy the given conditions :

Write the equations for the x- and y-axis.##### Answer:

Equation of the x-axis is y = 0.

Equation of the y-axis is x = 0

##### Question 47:

find the equation of the line, which satisfy the given conditions :

##### Answer:

##### Question 48:

find the equation of the line, which satisfy the given conditions :

Passing through (0, 0) with slope m.##### Answer:

##### Question 49:

find the equation of the line, which satisfy the given conditions :

##### Answer:

##### Question 50:

find the equation of the line, which satisfy the given conditions :

Intersecting the x-axis at a distance of 3 units to the left of origin with slope -2.##### Answer:

##### Question 51:

find the equation of the line, which satisfy the
given conditions :

Intersecting the y-axis at a distance of 2 units above
the origin and making an angle of 30$\xb0$ with positive
direction of the x-axis.

##### Answer:

##### Question 52:

find the equation of the line, which satisfy the
given conditions :

Passing through the points (–1, 1) and (2, –4).

##### Answer:

##### Question 53:

find the equation of the line, which satisfy the
given conditions :

Perpendicular distance from the origin is 5 units and
the angle made by the perpendicular with the positive
x-axis is 30$\xb0$.

##### Answer:

##### Question 54:

The vertices of a triangle PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find the equation of the median through the vertex R.

##### Answer:

##### Question 55:

Find the equation of the line through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).

##### Answer:

##### Question 56:

A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1 : n. Find the equation of the line.

##### Answer:

##### Question 57:

Find the equation of a line that cuts off equal intercepts on the co-ordinate axes and passes through the point (2, 3).

##### Answer:

##### Question 58:

Find the equation of a straight line passing through the point (2, 2) such that the sum of its intercepts on the axes is 9.

##### Answer:

##### Question 59:

##### Answer:

##### Question 60:

The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line.

##### Answer:

Since (–2, 9) lies on it, $\therefore $ 2 (–2) – 9(9) + k = 0 $\Rightarrow $ k = 85. Putting in (2), we get : 2x – 9y + 85 = 0, which is the reqd. equation of AB.##### Question 61:

The length L (in centimetres) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124·942 when C = 20 and L = 125·134 when C = 110, express L in terms of C.

##### Answer:

Assuming celsius (C) along the x-axis and length
L along the y-axis, we have the relation :

L = m C + k ...(1) [Form : y = mx + c]

When C = 20, L = 124·942

$\therefore $ 124·942 = m (20) + k ...(2)

When C = 110, L = 125·134

$\therefore $ 125·134 = m (110) + k ...(3)

Subtracting (2) from (3),

0·192 = 90 m

Putting in (3),

125·134 = 110 (0·213) + k

$\Rightarrow $ k = 125·134 – 23·430

= 101·704.

Putting in (1),

L = 0·213 C + 101·704,

which expresses L in terms of C.

##### Question 62:

##### Answer:

##### Question 63:

##### Answer:

##### Question 64:

Point R (h, k) divides a line segment between the axes in the ratio 1 : 2. Find the equation of the line.

##### Answer:

##### Question 65:

By using the concept of the equation of a line, prove that three points (3, 0), (–2, –2) and (8, 2) are collinear.

##### Answer:

##### Question 66:

Reduce the following equations into slope-intercept
form and find their slopes and the y-intercepts :

x + 7y = 0

##### Answer:

##### Question 67:

Reduce the following equations into slope-intercept
form and find their slopes and the y-intercepts :

6x + 3y – 5 = 0

##### Answer:

##### Question 68:

Reduce the following equations into slope-intercept
form and find their slopes and the y-intercepts :

y = 0

##### Answer:

The given equation is y = 0

$\Rightarrow $ y = 0.x + 0 [Form : y = mx + c]

Slope = 0 and y-intercept = 0.

##### Question 69:

Reduce the following equations into intercept form and
find their intercepts on the axes :

3x + 2y – 12 = 0

##### Answer:

##### Question 70:

Reduce the following equations into intercept form and
find their intercepts on the axes :

4x – 3y = 6

##### Answer:

##### Question 71:

Reduce the following equations into intercept form and
find their intercepts on the axes :

3y + 2 = 0

##### Answer:

##### Question 72:

Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive x-axis.

##### Answer:

##### Question 73:

Reduce the following equations into normal form. Find
their perpendicular distance from the origin and angle
between perpendicular and the positive x-axis.

y – 2 = 0

##### Answer:

##### Question 74:

Reduce the following equations into normal form. Find
their perpendicular distance from the origin and angle
between perpendicular and the positive x-axis.

x – y = 4

##### Answer:

##### Question 75:

Find the distance of the point (– 1, 1) from the
line :

12 (x + 6) = 5 (y – 2).

##### Answer:

The given line is :

12 (x + 6) = 5 (y – 2)

$\Rightarrow $ 12x + 72 = 5y – 10

$\Rightarrow $ 12x – 5y + 82 = 0 ....(1)

$\therefore $ Distance of (– 1, 1) from line (1)

##### Question 76:

##### Answer:

##### Question 77:

Find the distance between the parallel lines :

15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

##### Answer:

##### Question 78:

Find the distance between the parallel lines :

l (x + y) + p = 0 and l (x + y) – r = 0

##### Answer:

##### Question 79:

Find the equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (– 2, 3).

##### Answer:

Any line parallel to 3x – 4y + 2 = 0 ...(1)

is 3x – 4y + k = 0 ...(2)

Since (2) passes thro’ (– 2, 3),

$\therefore $ 3 (– 2) – 4 (3) + k = 0 $\Rightarrow $ k = 6 + 12 = 18.

Putting in (2), 3x – 4y + 18 = 0,

which is the reqd. equation.

##### Question 80:

Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x-intercept 3.

##### Answer:

The given line is x – 7y + 5 = 0 ...(1)

Since the x-intercept is 3,

$\therefore $ the reqd. line passes thro’ (3, 0).

Any line perpendicular to (1) is :

7x + y + k = 0 ...(2)

Since (2) passes thro’ (3, 0),

$\therefore $ 7 (3) + 0 + k = 0 $\Rightarrow $ k = – 21.

Putting is (2), 7x + y – 21 = 0,

which is the reqd. equation.

##### Question 81:

##### Answer:

##### Question 82:

The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0 at right angle. Find the value of h.

##### Answer:

##### Question 83:

##### Answer:

##### Question 84:

Two lines passing through the point (2, 3) intersect each other at an angle of 60°. If the slope of one line is 2, find the equation of the other line

##### Answer:

##### Question 85:

Find the equation of right-bisector of segment joining (3, 4) and (– 1, 2).

##### Answer:

##### Question 86:

Find the co-ordinates of the foot of perpendicular from a point (– 1, 3) to the line 3x – 4y – 16 = 0.

##### Answer:

##### Question 87:

The perpendicular from the origin to the line y = mx + c meets it at the point (– 1, 2). Find the values of m and c.

##### Answer:

##### Question 88:

##### Answer:

##### Question 89:

In the triangle ABC with vertices A (2, 3), B (4, – 1) and C (1, 2), find the equation and length of altitude from the vertex A.