10 Most SAQ’s of The Straight Line Chapter in Inter 1st Year Maths-1B (TS/AP)
4 Marks
SAQ-1 : Find the value of k, if the lines 2x – 3y + k = 0, 3x – 4y – 13 = 0, 8x – 11y – 33 = 0 are concurrent
Given lines:
$$3x – 4y – 13 = 0$$
$$8x – 11y – 33 = 0$$
Solving (1) and (2), we get P:
$$\frac{x}{(-4)(-33) – (-11)(-13)} = \frac{y}{-13(8) – (-33)(3)} = \frac{1}{3(-11) – 8(-4)}$$
$$\Rightarrow \frac{x}{132 – 143} = \frac{y}{-104 + 99} = \frac{1}{-33 + 32}$$
$$\Rightarrow \frac{x}{-11} = \frac{y}{-5} = \frac{1}{-1}$$
$$\Rightarrow x = -11 / -1 = 11; \, y = -5 / -1 = 5$$
Point of intersection is P(11,5)
Given lines are concurrent.
So, P(11,5) on $$2x – 3y + k = 0$$
$$\Rightarrow 2(11) – 3(5) + k = 0$$
$$\Rightarrow 22 – 15 + k = 0$$
$$\Rightarrow 7 + k = 0$$
$$\Rightarrow k = -7$$
Hence, the value of k = -7
SAQ-2 : Find the value of P if the lines 3x + 4y = 5, 2x + 3y = 4, px + 4y = 6 are concurrent
Given lines:
$$3x + 4y – 5 = 0$$
$$2x + 3y – 4 = 0$$
Solving (1) & (2), we get P:
$$\frac{x}{4(-4)-3(-5)} = \frac{y}{2(-5)-3(-4)} = \frac{1}{3(3)-2(4)}$$
$$\Rightarrow \frac{x}{-16 + 15} = \frac{y}{-10 + 12} = \frac{1}{9 – 8}$$
$$\Rightarrow \frac{x}{-1} = \frac{y}{2} = \frac{1}{1}$$
$$\Rightarrow x = -1/1 = -1; \, y = 2/2 = 2$$
Point of intersection is P(-1,2)
Given lines are concurrent.
So P(-1,2) lies on $$px + 4y = 6$$
$$\Rightarrow p(-1) + 4(2) = 6$$
$$\Rightarrow -p + 8 = 6$$
$$\Rightarrow p = 8 – 6 = 2$$
Hence, the value of p = 2
SAQ-3 : Find the equation of the straight line parallel to the line 3x + 4y = 7 and passing through the point of intersection of the lines x – 2y – 3 = 0, x + 3y – 6 = 0
Given the line is $$3x + 4y = 7$$ its slope $$m = -\frac{a}{b} = -\frac{3}{4}$$
Given equations:
$$x – 2y – 3 = 0$$
$$x + 3y – 6 = 0$$
Solving (1) & (2), we get P:
$$\frac{x}{(-2)(-6)-3(-3)} = \frac{y}{(-3)(1)-(-6)1} = \frac{1}{1(3)-1(-2)}$$
$$\Rightarrow \frac{x}{12 + 9} = \frac{y}{-3 + 6} = \frac{1}{3+2}$$
$$\Rightarrow \frac{x}{21} = \frac{y}{3} = \frac{1}{5}$$
$$\Rightarrow x = \frac{21}{5}, y = \frac{3}{5}$$
Point of intersection is $$P\left(\frac{21}{5}, \frac{3}{5}\right)$$
Line through $$P\left(\frac{21}{5}, \frac{3}{5}\right)$$ with slope $$-\frac{3}{4}$$ is given by
$$y – y_1 = m(x – x_1)$$
$$\Rightarrow y – \frac{3}{5} = -\frac{3}{4}(x – \frac{21}{5})$$
$$\Rightarrow 5y – 3 = -\frac{3}{4}(5x – 21)$$
$$\Rightarrow 4(5y – 3) = -3(5x – 21)$$
$$\Rightarrow 20y – 12 = -15x + 63$$
$$\Rightarrow 15x + 20y – 12 – 63 = 0$$
$$\Rightarrow 15x + 20y – 75 = 0$$
$$\Rightarrow 5(3x + 4y – 15) = 0$$
$$\Rightarrow 3x + 4y – 15 = 0$$
SAQ-4 : Find the equation of the line passing through the point of intersection of 2x + 3y = 1, 3x + 4y = 6 and perpendicular to the line 5x – 2y = 7
Given line is $$5x – 2y – 7 = 0$$
Its slope $$m = -\frac{a}{b} = -\frac{5}{-2} = \frac{5}{2}$$
The slope of its perpendicular is $$-\frac{1}{m} = \frac{2}{5}$$
Given:
$$2x + 3y – 1 = 0$$
$$3x + 4y – 6 = 0$$
Solving (1) & (2), we get P:
$$\frac{x}{3(-6)-4(-1)} = \frac{y}{3(-1)-2(-6)} = \frac{1}{2(4)-3(3)}$$
$$\Rightarrow \frac{x}{-18 + 4} = \frac{y}{-3 + 12} = \frac{1}{8 – 9}$$
$$\Rightarrow \frac{x}{-14} = \frac{y}{9} = -1$$
$$\Rightarrow x = 14, y = -9$$
Point of intersection is P = (14, -9)
A line perpendicular through P = (14,-9) with slope $$-\frac{2}{5}$$ is given by $$y – y_1 = -\frac{1}{m}(x – x_1)$$
$$\Rightarrow y + 9 = -\frac{2}{5}(x – 14)$$
$$\Rightarrow 5(y + 9) = -2(x – 14)$$
$$\Rightarrow 5y + 45 = -2x + 28$$
$$\Rightarrow 2x + 5y + 17 = 0$$
SAQ-5 : Find the value of k if the angle between the straight lines 4x – y + 7 = 0, kx – 5y – 9 = 0 is 450
Given line is $$4x – y + 7 = 0$$ with slope
$$m_1 = -\frac{a}{b} = -\frac{4}{-1} = 4$$
Another line is $$kx – 5y – 9 = 0$$ with slope
$$m_2 = -\frac{a}{b} = -\frac{k}{-5} = \frac{k}{5}$$
Given that the angle between the lines is 45∘, then
$$\tan \theta = \left| \frac{m_1 – m_2}{1 + m_1m_2} \right|$$
Substituting $$\tan 45^\circ = 1$$ we have:
$$1 = \left| \frac{4 – (k/5)}{1 + 4(k/5)} \right|$$
$$1 = \left| \frac{20 – k}{5 + 4k} \right|$$
Setting up the equation, we get two cases:
$$5 + 4k = 20 – k$$
$$5 + 4k = -(20 – k)$$
Solving the first case:
$$5 + 4k = 20 – k$$
$$5k = 15$$
$$k = 3$$
Solving the second case:
$$5 + 4k = -(20 – k)$$
$$5 + 4k = -20 + k$$
$$3k = -25$$
$$k = -\frac{25}{3}$$
Thus $$k = 3 or k = -\frac{25}{3}$$
SAQ-6 : A straight line with slope 1 passes through Q(-3,5) meets the line x + y – 6 = 0 at p. Find the distance PQ
Given slope m = 1
implies $$\tan \theta = \tan 45^\circ$$
so $$\theta = 45^\circ$$
Given point $$Q(x_1,y_1) = (-3,5)$$ and distance PQ = r then
$$p = \left(x_1 + r\cos \theta, y_1 + r\sin \theta\right) = \left(-3 + r\cos 45^\circ, 5 + r\sin 45^\circ\right)$$
$$= \left(-3 + r\left(\frac{1}{\sqrt{2}}\right), 5 + r\left(\frac{1}{\sqrt{2}}\right)\right)$$
$$= P\left(-3 + \frac{r}{\sqrt{2}}, 5 + \frac{r}{\sqrt{2}}\right)$$
But P lies on the line x + y – 6 = 0
$$⇒ x + y = 6$$
$$⇒ \left(-3 + \frac{r}{\sqrt{2}}\right) + \left(5 + \frac{r}{\sqrt{2}}\right) = 6$$
$$⇒ 2 + \frac{2r}{\sqrt{2}} = 6$$
$$⇒ \frac{2r}{\sqrt{2}} = 4$$
$$⇒ \frac{r}{\sqrt{2}} = 2$$
$$⇒ r = 2\sqrt{2}$$
Hence, distance $$PQ = 2\sqrt{2}$$
SAQ-7 : A straight line through Q(√3,2) makes an angle π/6 with the positive direction of the X-axis. If the straight line intersects the line√3x – 4y + 8 = 0 at p find the distance PQ
Given angle $$\theta = \frac{\pi}{6} = 30^\circ$$
Given point $$Q(x_1,y_1) = (\sqrt{3},2)$$ and distance PQ = r then
$$P = (x_1 + r\cos\theta, y_1 + r\sin\theta)$$
$$= \left(\sqrt{3} + r\cos 30^\circ, 2 + r\sin 30^\circ\right)$$
$$= \left(\sqrt{3} + r\left(\frac{\sqrt{3}}{2}\right), 2 + r\left(\frac{1}{2}\right)\right)$$
$$= P\left(\sqrt{3} + \frac{\sqrt{3}r}{2}, 2 + \frac{r}{2}\right)$$
But, P lies on the line $$\sqrt{3}x – 4y + 8 = 0$$
$$⇒ \sqrt{3}\left(\sqrt{3} + \frac{\sqrt{3}r}{2}\right) – 4\left(2 + \frac{r}{2}\right) + 8 = 0$$
$$⇒ 3 + \frac{3r}{2} – 8 – 2r + 8 = 0$$
$$⇒ \frac{3r}{2} – 2r = 0$$
$$⇒ \frac{3r – 4r}{2} = 0$$
$$⇒ -\frac{r}{2} = 0$$
$$⇒ 3 + \frac{3r}{2} – 8 – 2r + 8 = 0$$
Hence, the distance
$$PQ = r = 6$$
SAQ-8 : Find the points on the line 3x – 4y – 1 = 0 which are at a distance of 5 units from the point (3,2)
Given the line is $$3x – 4y – 1 = 0$$
Its slope $$m = -\frac{a}{b} = -\frac{3}{-4} = \frac{3}{4}$$
$$m = \tan \theta = \frac{3}{4}$$
$$\Rightarrow \cos \theta = \frac{4}{5}, \sin \theta = \frac{3}{5}$$
Given the distance $$|r| = 5$$
Given point $$(x_1,y_1) = (3,2)$$
The required points $$(x,y) = (x_1 \pm r\cos \theta, y_1 \pm r\sin \theta) = (3 \pm 5\cdot\frac{4}{5}, 2 \pm 5\cdot\frac{3}{5}) = (3 \pm 4, 2 \pm 3)$$
$$= (3 + 4, 2 + 3) = (7,5) or (3 – 4, 2 – 3) = (-1, -1)$$
The required points are (7,5) and (-1,-1)
SAQ-9 : If the straight lines ax + by + c = 0, bx + cy + a = 0 and cx + ay + b = 0 are concurrent then prove that a3 + b3 + c3 = 3abc
Given lines:
$$ax + by + c = 0$$
$$bx + cy + a = 0$$
Solving (1) and (2), we aim to find the point of intersection P.
$$\frac{x}{ba – c^2} = \frac{y}{cb – a^2} = \frac{1}{ac – b^2}$$
After rearranging, we find expressions for x and y as follows:
$$x = \frac{ab – c^2}{ca – b^2} \quad \text{and} \quad y = \frac{bc – a^2}{ca – b^2}$$
Thus, the point of intersection P is given by $$P = \left( \frac{ab – c^2}{ca – b^2}, \frac{bc – a^2}{ca – b^2} \right)$$
But P lies on the line cx + ay + b = 0 leading to the equation:
$$c\left(\frac{ab – c^2}{ca – b^2}\right) + a\left(\frac{bc – a^2}{ca – b^2}\right) + b = 0$$
Multiplying throughout by $$ca – b^2$$ to clear the denominators, we get:
$$c(ab – c^2) + a(bc – a^2) + b(ca – b^2) = 0$$
Expanding, we have:
$$cab – c^3 + abc – a^3 + bca – b^3 = 0$$
Since cab, abc, and bca are all the same term, we can combine them:
$$3abc = a^3 + b^3 + c^3$$
Hence, it is proved that $$a^3 + b^3 + c^3 = 3abc$$ under the given conditions.
SAQ-10 : Transform the equation x/a + y/b = 1 into normal form. If the perpendicular distance of the straight line from the origin is p then deduce that 1/p2 = 1/a2 + 1/b2
The given equation is $$x/a + y/b = 1$$ which can be rewritten as:
$$bx + ay = ab$$
Dividing both sides by $$\sqrt{a^2 + b^2}$$ gives
$$\frac{bx}{\sqrt{a^2 + b^2}} + \frac{ay}{\sqrt{a^2 + b^2}} = \frac{ab}{\sqrt{a^2 + b^2}}$$
This equation resembles the normal form of a line, $$x\cos \alpha + y\sin \alpha = p$$ where
$$\cos \alpha = \frac{b}{\sqrt{a^2 + b^2}}, \sin \alpha = \frac{a}{\sqrt{a^2 + b^2}}, \text{ and } p = \frac{ab}{\sqrt{a^2 + b^2}}$$
Thus, we identify $$p = \frac{ab}{\sqrt{a^2 + b^2}}$$
Squaring both sides of the expression for p, we get:
$$p^2 = \frac{a^2b^2}{a^2 + b^2}$$
Rearranging for $$1/p^2$$ we have:
$$1/p^2 = \frac{a^2 + b^2}{a^2b^2} = \frac{a^2}{a^2b^2} + \frac{b^2}{a^2b^2} = \frac{1}{b^2} + \frac{1}{a^2}$$
Hence, it’s proven that $$1/p^2 = 1/a^2 + 1/b^2$$