Straight Lines Important Questions

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Straight lines Important Questions covers important topics like General equation of a line, the slope of a line and the distance of a point from a line, and various forms of the equation of a line. A Straight Line in mathematics is defined as a structure that has no curvature but has negligible width and depth. It is essentially formed by connecting two points.

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Straight Lines Detailed Video Explanation:

Very Short Answer Questions [1 Marks Question]

Ques. What is the equation of the line through (-2,3 ) with slope -4?

Ans. According to the question,

m = -4

Given points ( x₀,y₀ ) is ( -2,3 )

As we know the Slope-Intercept Form Formula is

y-y₀ = m(x-x₀)

y - 3 = -4(x+2)

4x + y + 5 = 0

Hence the desired equation is 4x+ y + 5 = 0

Ques. What is the equation of the line, which makes intercepts on the x and y as -3 and 2 respectively?

Ans. Here

a = -3

b = 2

We know that the equation of the line making intercepts a and b on the x and y axis is

xa + yb = 1

Thus by substitution

x.(-3) + y.2 = 1

2y - 3x -1 =0

Ques. What is the equation of a straight line parallel to the y-axis and passing through the point (4,−2)?

Ans. Given the point is

(4,−2)

Equation of line parallel to y-axis is

x = a...(i)

Substituting equation (i) passing through(4,−2)

a = 4

So,

x = 4

x − 4 = 0

Hence, the equation is x - 4 = 0

Ques. The equation of the sides of a triangle is x−3y=0. 4x+3y=5 and 3x+y=0. Where do the lines 3x−4y=0 pass through?

Ans. Two sides x − 3y = 0 and 3x + y = 0 are perpendicular to each other.

Its orthocentre is the point of intersection of x − 3y = 0 i.e. (0, 0) so,

the line 3x−4y = 0 passes through the circumference of the triangle.

Short Answer Questions [2 Marks Question]

Ques. What is the distance between the lines 5x+3y−7=0 and 15x+9y+14=0?

Ans. The given lines can be written as -

5x + 3y - 7 = 0 — 1)

5x + 3y + 14/3 = 0 —- 2)

Let d be the distance between the lines 5x + 3y - 7 = 0 and 15 x + 9y + 14 = 0

Then d = -7-14352+ 32

d = \(35 \over 3\sqrt34\)

Ques. (-4, -5) is one vertex and 7x − y + 8 = 0 is one diagonal of a square. What is the equation of the second diagonal?

Ans. Equation of perpendicular line to

7x − y + 8 = 0 is x + 7y = h

x + 7y = h which it passes through (−4, 5)

h = 31

So, the equation of another diagonal is x + 7y = 31

Ques. What is the equation of the straight line which is perpendicular to y=x and passes through (3, 2)?

Ans. The slope of the given line y = x is 1

Therefore, the slope of the required line which is perpendicular to the given line is -1

Required line is: y = −x + c

It passes through (3,2)

\(⇒\)2 = −3 +C

\(⇒\)C=5

∴ Required line:

y = −x + 5

\(⇒\) x + y = 5

Short Answer Questions [3 Marks Question]

Ques. What is the slope of a line that passes through the origin, and the mid-point of the segment joining the points P (0, -4) and B (8, 0)?

Ans. Given,

Coordinates of the mid-point of the line segment joining the points P (0, -4) and B (8, 0) are:

[(0+8)/2, (-4+0)/2] = (4, -2)

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

m = (y2 -y1)/(x2 -x1), where (x2 is not equal to x1)

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

m = (-2-0)/(4-0)

m = -2/4

m = -1/2

Hence, the required slope of the line is -1/2

Ques. What is the equation of the line which is at a perpendicular distance of 5 units from the origin and the angle made by the perpendicular with the positive x-axis is 30°?

Ans. If p is the length of the normal from the origin to a line

ω is the angle made by the normal with the positive direction of the x-axis

Then, the equation of the line for the given condition is written by

x cos\( ω\) + 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

x(\(√3/2\)) + y(\(½\)) = 5

It becomes

\(√ 3\)x + y = 10

Thus, the required equation of a line is \(√ 3\)x + y = 10

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

Ans. The equation of the line is given as x – 7y + 5 = 0.

The above equation can be written in the form y = mx+c

Thus, the above equation is written as:

y = (1/7)x + (5/7)

From the above equation, we can say that,

The slope of the line perpendicular to the given line is

m = -1/(1/7) = -7

Hence, the equation of a line with slope -7 and intercept 3 is given as:

y = m (x – d)

\(⇒\) y= -7(x-3)

\(⇒\) y=-7x + 21

7x+ y = 21

Hence, the equation of a line that is perpendicular to the line x – 7y + 5 = 0 with x-intercept 3 is 7x+ y = 21.

Very Long Answer Questions [5 Marks Question]

Ques. 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.

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

From the given condition, the perpendicular from the origin meets the given line at (-1, 2).

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

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

= 2/-1 = -2

Therefore,

m (– 2) = -1 (Since the two lines are perpendicular)

m= \( ½\)

Since points (-1, 2) lie on the given line, it satisfies the equation y = MX + c.

Now, substitute the value of m, (x, y) coordinates in the equation:

2 = m(-1) + c

2 =\( ½\)(-1) + c

2 = -\( ½\) + c

C = 2 + (\( ½\))

C = 5/2

Therefore, the values of m and c are 1/2 and 5/2 respectively.

Ques. What are the points on the x-axis whose distance from the line equation (x/3) + (y/4) = 1 is given as 4 units.

Ans. Given that,

The equation of a line = (x/3) + (y/4) = 1

It can be written as:

4x + 3y -12 = 0 …(1)

Compare the equation (1) with the general line equation Ax + By + C = 0,

we get the values 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.

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

D = |Ax1+By1 + \(C|/√A2 \)+ B2

Now, substitute the values in the above formula, we get:

4 = |4a+0 + -12|/ \(√42\) + 32

\(⇒\)4 = |4a-12|/5

\(⇒\)|4a-12| = 20

\(⇒±\) (4a-12)= 20

\(⇒\) (4a-12)= 20 or -(4a-12) =20

Therefore, it can be written as:

(4a-12)= 20

4a = 20+12

4a = 32

a = 8

(or)

-(4a-12) =20

-4a +12 =20

-4a = 20-12

-4a= 8

a= -2

⇒ a= 8 or -2

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

Ques. What is the equation of the line that has y−intercept 4 and is perpendicular to the line y=3x−2? Also, find the equation of the line, which makes intercepts −3 and 2 on the y-axis respectively.

Ans. (a) The given equation of is y=3x−2

Express the given equation as slope-intercept form y=mx+c

where,

(Slope)m = coefficient of x

m1 = 3

When the lines are perpendicular. Then the product of the slope is −1

∴m1.m2 =−1

3.m2=−1

m2=−13

Given, the y−intercept of the other line is 4.

Therefore, the required equation of the line using the slope-intercept form y=mx+c

y = −13x+4

(b.) Given,(x−intercept) a =−3

(y−intercept) b =2

The required equation is given by xa+yb=1

a=−3,

b=2

∴x−3+y2=1

2x−3y+6=0

Maths Related Links:

Horizontal and vertical lines	Analytical Geometry	Slope Formula
Slope Intercept Form Formula	Point Gradient Formula	Perpendicular line formula
Point slope form formula	Different forms of the equation of line	Slope intercept form

Maths Study Guides:

Math Formula	Difference between in Maths	Algebra
Arithmetic	Geometry	Mensuration

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