Straight Lines Important Questions

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. 

The video below explains this:

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:

Maths Study Guides: