The Straight Line (VSAQs)
Maths-1B | 3. The Straight Line – VSAQs:
Welcome to VSAQs in Chapter 3: The Straight Line. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
VSAQ-1 : Find the value of x, if the slope of the line passing through (2,5),(x,3) is 2
We have points A = (2,5) and B = (x,3)
The slope of line AB is given by:
$$m_1 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{3 – 5}{x – 2} = \frac{-2}{x – 2}$$
Given that the slope $$m_2 = 2$$ and setting $$m_1 = m_2$$ we get:
$$\frac{-2}{x – 2} = 2$$
Solving for x, we have:
$$x – 2 = \frac{-2}{2} = -1$$
$$x = -1 + 2 = 1$$
Therefore $$x = 1$$
VSAQ-2 : Find the value of y, if the line joining (3,y), (2,7) is parallel to the line joining the points (-1,4), (0,6) is 2
We have points A = (3,y) and B = (2,7)
The slope of line AB is
$$m_1 = \frac{y_2 – y_1}{x_2 – x_1} = \frac{7 – y}{2 – 3} = \frac{7 – y}{-1} = y – 7$$
We also have points C = (-1,4) and D = (0,6)
The slope of line CD is
$$m_2 = \frac{6 – 4}{0 + 1} = \frac{2}{1} = 2$$
Given that $$m_1 = m_2$$ we get
$$y – 7 = 2$$
Solving for y, we find:
$$y = 2 + 7 = 9$$
Thus $$y = 9$$
VSAQ-3 : Find the equation of the straight line passing through (-4,5) and cutting off equal intercepts on the coordinate axes
Given that the intercepts are equal, we assume the intercepts to be a, a
The intercept form of the equation is $$x/a + y/b = 1$$ Since a = b we have:
$$x/a + y/a = 1$$
Multiplying through by a gives:
$$x + y = a$$
Given that the point (-4,5) lies on this line:
$$-4 + 5 = a$$
$$a = 1$$
Thus, the equation becomes:
$$x + y = 1$$
Or equivalently:
$$x + y – 1 = 0$$
VSAQ-4 : Find the equation of the straight line passing through the point (-2,4) and making intercepts whose sum is zero
Given that the sum of intercepts is zero, we assume the intercepts to be a, -a
The intercept form of the equation is $$x/a + y/b = 1$$ Given b = -a we have:
$$x/a + y/(-a) = 1$$
Simplifying gives:
$$x – y = a$$
Given that the point (-2,4) lies on this line:
$$-2 – 4 = a$$
$$a = -6$$
Thus, the equation becomes:
$$x – y = -6$$
Or equivalently:
$$x – y + 6 = 0$$
VSAQ-5 : Find the equation of the straight line passing through the point (2,3) and making intercepts, whose sum is zero
Given that the sum of intercepts is zero, we take the intercepts as a, -a
In intercept form: $$x/a + y/b = 1$$
With b = -a it simplifies to:
$$x/a + y/(-a) = 1$$
Which further simplifies to:
$$x – y = a$$
Given that the point (2, 3) lies on this line:
$$2 – 3 = a$$
Thus $$a = -1$$
Substituting a = -1 into equation (1):
$$x – y = -1$$
$$x – y + 1 = 0$$
VSAQ-6 : Transform the equation 4x – 3y + 12 = 0 into (i) slope intercept form (ii) intercept form
(i) Slope intercept form is $$y = mx + c$$
$$4x – 3y + 12 = 0$$
$$3y = 4x + 12$$
$$y = \frac{4}{3}x + 4$$
$$y = \frac{4}{3}x + 4$$
(ii) Intercept form is $$x/a + y/b = 1$$
$$4x – 3y + 12 = 0$$
$$4x – 3y = -12$$
$$\frac{4x}{-12} + \frac{-3y}{-12} = 1$$
$$\frac{x}{-3} + \frac{y}{4} = 1$$
$$x/-3 + y/4 = 1$$
VSAQ-7 : Transform the equation √3x + y = 4 into (i) Slope intercept form (ii) intercept form
(i) Slope intercept form is y = mx + c
$$\sqrt{3}x + y = 4$$
$$y = -\sqrt{3}x + 4$$
$$y = -\sqrt{3}x + 4$$
(ii) Intercept form is x/a + y/b = 1
$$\sqrt{3}x + y = 4$$
$$\frac{\sqrt{3}x}{4} + \frac{y}{4} = 1$$
$$\frac{x}{4/\sqrt{3}} + \frac{y}{4} = 1$$
$$x/(4/\sqrt{3}) + y/4 = 1$$
VSAQ-8 : Transform the equation x + y + 1 = 0 into normal form
The normal form of a line equation is given as $$x\cos\alpha + y\sin\alpha = p$$ where $$p > 0$$
$$x + y + 1 = 0$$
$$x + y = -1$$
$$-x – y = 1$$
Next, divide through by $$\sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$$
$$\frac{-x}{\sqrt{2}} – \frac{-y}{\sqrt{2}} = \frac{1}{\sqrt{2}}$$
$$x\left(-\frac{1}{\sqrt{2}}\right) + y\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{\sqrt{2}}$$
$$x\cos 225^\circ + y\sin 225^\circ = \frac{1}{\sqrt{2}}$$
Therefore, this is the normal form of the equation of the line.
VSAQ- 9 : Transform the equation 3x + 4y + 12 = 0 into Normal form
Normal form is $$x \cos \alpha + y \sin \alpha = p p>0$$
$$3x + 4y + 12 = 0$$
$$3x + 4y = -12$$
$$-3x – 4y = 12$$
Now dividing $$\sqrt{(-3)^2 + (-4)^2}$$
$$= \sqrt{9 + 16} = \sqrt{25} = 5$$
$$-\frac{3}{5}x – \frac{4}{5}y = \frac{12}{5}$$
We write it as $$x \cos \alpha + y \sin \alpha = p$$
Where $$\cos \alpha = -\frac{3}{5} \sin \alpha = -\frac{4}{5} p = \frac{12}{5}$$
Hence $$\tan \alpha = \frac{4}{3}$$
$$\alpha = \tan^{-1}\left(\frac{4}{3}\right)$$
VSAQ-10 : Find the equation of the straight line parallel to the line 2x + 3y + 7 = 0 and passing through the point (5,4)
Given line is $$2x + 3y + 7 = 0$$
Its slope $$m = -\frac{a}{b} = -\frac{2}{3}$$
Parallel line through (5,4) with slope $$-\frac{2}{3}$$ is $$y – y_1 = m(x – x_1)$$
$$y – 4 = -\frac{2}{3}(x – 5)$$
$$3(y – 4) = -2(x – 5)$$
$$3y – 12 = -2x + 10$$
$$2x + 3y – 22 = 0$$
VSAQ-11 : Find the equation of the straight line perpendicular to the line 5x – 3y + 1 = 0 and passing through the point (4,-3)
Given line is $$5x – 3y + 1 = 0$$
Its slope $$m = -\frac{a}{b} = -\frac{5}{3} = \frac{5}{3}$$
Slope of its perpendicular is $$-\frac{1}{m} = -\frac{3}{5}$$
Perpendicular through (4,-3) with slope $$-\frac{3}{5}$$ is $$y – y_1 = -\frac{1}{m}(x – x_1)$$
$$y + 3 = -\frac{3}{5}(x – 4)$$
$$5(y + 3) = -3(x – 4)$$
$$5y + 15 = -3x + 12$$
$$3x + 5y + 3 = 0$$
VSAQ-12 : Find the area of the triangle formed by the line 3x – 4y + 12 = 0 with the coordinate axes
Given line is $$3x – 4y + 12 = 0$$
On comparing with $$ax + by + c = 0$$ we get a = 3 b = -4 c = 12
Area of the triangle is
$$\Delta = \frac{c^2}{2|ab|} = \frac{12^2}{2|(3)(-4)|} = \frac{144}{24} = 6$$
Hence area of the triangle is 6 sq.units.
VSAQ-13 : Find the value of a if the area of the triangle formed by the lines x = 0, y = 0, 3x + 4y = a is 6 sq. units
Given line is $$3x + 4y = a$$
$$3x + 4y – a = 0$$
On comparing with $$ax + by + c = 0$$ we get a = 3 b = 4 c = -a
Area of the triangle
$$\Delta = \frac{c^2}{2|ab|} = \frac{(-a)^2}{2|(3)(4)|} = \frac{a^2}{24}$$
But given area is 6
$$\frac{a^2}{24} = 6$$
$$a^2 = 6 \times 24 = 144 = 12^2$$
$$a = \pm 12$$
Hence value of $$a = \pm 12$$
VSAQ-14 : Find the ratio in which the straight line 2x + 3y – 5 = 0 divides the line joining points (0,0) and (-2,1)
We take $$L = 2x + 3y – 5$$ and
$$A(x_1, y_1) = (0,0) B(x_2, y_2) = (-2,1)$$
$$L_{11} = 2(0) + 3(0) – 5 = -5$$
$$L_{22} = 2(-2) + 3(1) – 5 = -6$$
Required ratio is $$-L_{11} : L_{22}$$
$$= -[(-5) : (-6)] = -5 : 6$$
VSAQ-15 : Find the ratio in which the straight line 2x + 3y – 20 = 0 divides the line joining the points (2,3) and (2,10)
Let $$L = 2x + 3y – 20$$ and
$$A(x_1, y_1) = (2,3) B(x_2, y_2) = (2,10)$$
Now $$L_{11} = 2(2) + 3(3) – 20 = -7$$
$$L_{22} = 2(2) + 3(10) – 20 = 14$$
Required ratio is $$-L_{11} : L_{22}$$
$$= -(-7) : 14$$
$$= 7 : 14 = 1 : 2$$
VSAQ-16 : Find the distance between the parallel lines 5x – 3y – 4 = 0, 10x 6y – 9 = 0
We write the first line $$5x – 3y – 4 = 0 as 10x – 6y – 8 = 0$$
Second line is $$10x – 6y – 9 = 0$$
The distance between (1) & (2) is
$$\frac{|c_1 – c_2|}{\sqrt{a^2 + b^2}} = \frac{|-8 + 9|}{\sqrt{10^2 + 6^2}} = \frac{|1|}{\sqrt{100 + 36}} = \frac{1}{\sqrt{136}}$$
VSAQ-17 : Find the distance between the parallel lines 3x + 4y – 3 = 0 and 6x + 8y – 1 = 0
We write the first line $$3x + 4y – 3 = 0 as 6x + 8y – 6 = 0$$
Second line is $$6x + 8y – 1 = 0$$
The distance between (1) & (2) is
$$\frac{|c_1 – c_2|}{\sqrt{a^2 + b^2}} = \frac{|-6 + 1|}{\sqrt{6^2 + 8^2}} = \frac{|-5|}{\sqrt{36 + 64}} = \frac{5}{\sqrt{100}} = \frac{5}{10} = \frac{1}{2}$$
VSAQ-18 : Find the value of p, if the straight lines x + p = 0, y + 2 = 0, 3x + 2y + 5 = 0 are concurrent
Given line $$x + p = 0$$
$$x = -p$$
$$y + 2 = 0$$
point of intersection is $$(-p,-2)$$
But (-p,-2) lies on 3x + 2y + 5 = 0
$$3(-p) + 2(-2) + 5 = 0$$
$$-3p – 4 + 5 = 0$$
$$-3p + 1 = 0$$
$$-3p = -1$$
$$3p = 1$$
$$p = \frac{1}{3}$$