# Class 11 NCERT Solutions- Chapter 10 Straight Lines – Exercise 10.2 | Set 1

Last Updated : 05 Apr, 2021

### Question 1. Write the equations for the x-and y-axes.

Solution:

The y-coordinate of every point on x-axis is 0 and the x-coordinate of every point on y-axis is 0

So, the equation of x-axis is y = 0 and the equation of y-axis is y = 0.

### Question 2. Passing through the point (â€“ 4, 3) with slope 1/2.

Solution:

Given that point p(x1, y1) is (-4, 3) and slope m = 1/2

Equation of Line can be derived by formula y â€“ y1 = m (x â€“ x1),

Where m is slope of line and (x1, y1) is co-ordinate of p from which line passes.

y â€“ 3 = 1/2(x â€“ (-4))

y â€“ 3 = 1/2 (x + 4)

2(y â€“ 3) = x + 4

2y â€“ 6 = x + 4

x + 4 â€“ (2y â€“ 6) = 0

x + 4 â€“ 2y + 6 = 0

x â€“ 2y + 10 = 0

So, the equation of the line is x â€“ 2y + 10 = 0.

### Question 3. Passing through (0, 0) with slope m.

Solution:

Given that Point p(x1, y1) is (0, 0) and slope is m.

So, the equation of Line can be derived by formula y â€“ y1 = m (x â€“ x1),

Where m is slope of line and (x1, y1) is co-ordinate from which line passes.

So, y â€“ 0 = m (x â€“ 0)

y = mx

y â€“ mx = 0

So, the equation of the line is y â€“ mx = 0.

### Question 4. Passing through (2, 2âˆš3) and inclined with the x-axis at an angle of 75o.

Solution:

Given that point p(x1, y1) is (2, 2âˆš3) and Î¸ = 75Â°

So, the equation of line is (y â€“ y1) = m (x â€“ x1)

where, m = slope of line = tan Î¸ and (x1, y1) are the points through which line passes

So, m = tan 75Â°

Now, finding tan 75Â° using below formula:

tan(A + B) = (tanA + tanB)/(1 – tanA.tanB)

tan 75Â° = tan(45Â° + 30Â°)

tan 75Â° = (tan 45Â° + tan 30Â°)/(1 – tan 30Â°.tan 45Â°)

tan 75Â° = (1 + 1/âˆš3)/(1 – 1/âˆš3)

tan 75Â° = (âˆš3 + 1)/(âˆš3 – 1)

By rationalizing, we get

tan 75Â° = 2 + âˆš3

Equation of Line will be,

(y â€“ 2âˆš3) = (2 + âˆš3)(x â€“ 2)

y â€“ 2âˆš3 = 2 x â€“ 4 + âˆš3x â€“ 2 âˆš3

y = 2 x â€“ 4 + âˆš3x

(2 + âˆš3)x â€“ y â€“ 4 = 0

So, the equation of the line is (2 + âˆš3)x â€“ y â€“ 4 = 0.

### Question 5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope â€“2.

Solution:

Given that Slope m = â€“2, x-intercept = 3,

that means Line is passing from point p(x1, y1) that is (3, 0)

Equation of line will be

y â€“ 0 = (â€“2) Ã— (x âˆ’ 3).

y = (â€“2) Ã— (x + 3)

y = â€“2x â€“ 6

2x + y + 6 = 0

So, the equation of the line is 2x + y + 6 = 0.

### Question 6. Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30Â° with the positive direction of the x-axis.

Solution:

Given that Î¸ = 30Â° then slope(m) will be = tan Î¸

m = tan30Â° = (1/âˆš3)

As Y intercept is 2 that means line is passing from (0, 2) point p(x1, y1) = (0, 2)

Equation of Line will be

y â€“ 2 = (1/âˆš3)x

y = (1/âˆš3)x + 2

y = (x + 2âˆš3) / âˆš3

âˆš3 y = x + 2âˆš3

x â€“ âˆš3 y + 2âˆš3 = 0

So, the equation of the line is x â€“ âˆš3 y + 2âˆš3 = 0.

### Question 7. Passing through the points (â€“1, 1) and (2, â€“ 4).

Solution:

Given that point p1(x1, y1) is (â€“1, 1) and point p2(x2, y2) is (2, â€“4),

It is mention in question that line passes from p1 and p2.

That means, the Slope(m) of line will be (y2 â€“ y1)/(x2 â€“ x1)

m = (â€“4 â€“ 1)/(2 â€“ (â€“1))

m = â€“5/3

Equation of Line will be

(y â€“ y1) = m(x â€“ x1)

y â€“ 1 = â€“5/3 (x + 1)

3 (y â€“ 1) = (â€“5)(x + 1)

3y â€“ 3 = â€“5x â€“ 5

3y â€“ 3 + 5x + 5 = 0

5x + 3y + 2 = 0

So, the equation of the line is 5x + 3y + 2 = 0.

### Question 8. Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive x-axis is 30Â°.

Solution:

Given that p = 5 and Ï‰ = 30Â°

We know that the equation of the line having normal distance p from the origin and

angle Ï‰ which the normal makes with the positive direction of x-axis is given by x cos Ï‰ + y sin Ï‰ = p.

On substituting the values in the equation, we get

x cos30Â° + y sin30Â° = 5

x(âˆš3 / 2) + y(1/2) = 5

âˆš3 x + y = 5(2) = 10

âˆš3 x + y â€“ 10 = 0

So, the equation of the line is âˆš3 x + y â€“ 10 = 0.

### Question 9. The vertices of Î”PQR are P (2, 1), Q (â€“2, 3), and R (4, 5). Find the equation of the median through the vertex R.

Solution:

Given Vertices of Î”PQR that are P (2, 1), Q (â€“2, 3) and R (4, 5)

Let RS be the median of vertex R => S is a midpoint of PQ.

As S is midpoint of PQ => S = (P + Q)/2

S = (2 – 2, 1 + 3)/2

S = (0, 2)

Equation of the line passing through the points (x1, y1) and (x2, y2) is given by

y â€“ 5 = â€“3/ â€“4(x â€“ 4)

(â€“4)(y â€“ 5) = (â€“3)(x â€“ 4)

â€“4y + 20 = â€“3x + 12

â€“4y + 20 + 3x â€“ 12 = 0

3x â€“ 4y + 8 = 0

So, the equation of median through the vertex R is 3x â€“ 4y + 8 = 0.

### Question 10. Find the equation of the line passing through (â€“3, 5) and perpendicular to the line through the points (2, 5) and (â€“3, 6).

Solution:

Given that Points are (2, 5) and (-3, 6).

So, the slope, m1 = (y2 â€“ y1)/(x2 â€“ x1)

= (6 â€“ 5)/(â€“3 â€“ 2)

= 1/â€“5 = â€“1/5

As we know that two non-vertical lines are perpendicular to each other

if their slopes are negative reciprocals of each other.

Then, m = (â€“1/m1)

= â€“1/(â€“1/5)

= 5

As we know that the point p(x, y) lies on the line with

slope m through the fixed point (x1, y1),

If its coordinates satisfy the equation y â€“ y1 = m (x â€“ x1)

Then, y â€“ 5 = 5(x â€“ (â€“3))

y â€“ 5 = 5x + 15

5x + 15 â€“ y + 5 = 0

5x â€“ y + 20 = 0

So, the equation of the line is 5x â€“ y + 20 = 0

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