Maths Solve the Following Question.(3 Marks)

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is

given by u + kv = 0.

(x + y + 9) + k(2x + 3y + 1) = 0 ……(i)

⇒ x + y + 9 + 2kx + 3ky + k = 0

⇒ (1 + 2k)x + (1 + 3k)y + 9 + k = 0

But, x-intercept of this line is 1.

$\Rightarrow \frac{-(9+k)}{1+2 k}$

⇒ -9 – k = 1 + 2k

$\Rightarrow k=\frac{-10}{3}$

Substituting the value of k in (i), we get

$(x+y+9)+\left(\frac{-10}{3}\right)(2 x+3 y+1)=0$

⇒ 3(x + y + 9) – 10(2x + 3y + 1) = 0 ⇒ 3x + 3y + 27 – 20x – 30y – 10 = 0 ⇒ -17x – 27y+ 17 = 0 ⇒ 17x + 27y – 17 = 0, which is the equation of the required line.

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.

(x + y – 3) + k(2x – y + 1) = 0 …..(i)

x + y – 3 + 2kx – ky + k = 0

x + 2kx + y – ky – 3 + k = 0

(1 + 2k)x + (1 – k)y – 3 + k = 0

But, this line is parallel to X-axis

Its slope = 0

$\begin{aligned} & \Rightarrow \frac{-(1+2 k)}{1-k}=0 \\ & \Rightarrow 1+2 k=0 \\ & \Rightarrow k=\frac{-1}{2}\end{aligned}$

Substituting the value of k in (i), we get

$(x+y-3)+\frac{-1}{2}(2 x-y+1)=0$

⇒ 2(x + y – 3) – (2x – y + 1 ) = 0 ⇒ 2x + 2y – 6 – 2x + y – 1 = 0 ⇒ 3y – 7 = 0, which is the equation of the required line.

Slope of BC = -3

Slope of $A X=\frac{1}{3} \ldots \ldots[\because A X \perp B C]$

Since altitude AX passes through $(1,4)$ and has slope $\frac{1}{3}$.

equation of altitude AX is

$y-4=\frac{1}{3}(x-1)$

3y – 12 = x – 1 ∴ x – 3y + 11 = 0

Since both the points A and C have same x co-ordinates i.e. 1 the points A and C lie on the line x = 1. AC is parallel to Y-axis and therefore, altitude BY is parallel to X-axis. Since the altitude BY passes through B(2, 3), the equation of altitude BY is y = 3. Also, slope of AB = -1 Slope of CZ = 1 Since altitude CZ passes through (1, 6) and has slope 1, equation of altitude CZ is y – 6 = 1(x – 1) ∴ x – y + 5 = 0

Slope of perpendicular bisector of $B C$ is $\frac{1}{3}$ and the line passes through $\left(\frac{3}{2}, \frac{9}{2}\right)$.

Equation of the perpendicular bisector of side BC is

$\left(y-\frac{9}{2}\right)=\frac{1}{3}\left(x-\frac{3}{2}\right)$

3(2y – 9) = (2x – 3)

6y – 27 = 2x – 3

2x – 6y + 24 = 0

∴ x – 3y + 12 = 0

Since both the points A and C have same x co-ordinates i.e. 1 the points A and C lie on the line x = 1. AC is parallel to Y-axis and therefore, perpendicular bisector of side AC is parallel to X-axis. Since, the perpendicular bisector of side AC passes through E(1, 5). The equation of perpendicular bisector of side AC is y = 5.

Slope of side $A B=\left(\frac{3-4}{2-1}\right)=-1$

Slope of perpendicular bisector of $A B$ is 1 and the line passes through $\left(\frac{3}{2}, \frac{7}{2}\right)$.

Equation of the perpendicular bisector of side AB is

$\begin{aligned} & \left(y-\frac{7}{2}\right)=1\left(x-\frac{3}{2}\right) \\ & 2 y-7=2 x-3 \\ & 2 x-2 y+4=0 \\ & \therefore x-y+2=0\end{aligned}$

Then $\mathrm{D} \equiv\left(\frac{2+1}{2}, \frac{3+6}{2}\right)=\left(\frac{3}{2}, \frac{9}{2}\right)$

$\mathrm{E} \equiv\left(\frac{1+1}{2}, \frac{6+4}{2}\right)=(1,5)$

$\mathrm{F} \equiv\left(\frac{1+2}{2}, \frac{4+3}{2}\right)=\left(\frac{3}{2}, \frac{7}{2}\right)$

Equation of median $\mathrm{AD}$ is

$\begin{aligned} & \frac{y-4}{\frac{9}{2}-4}=\frac{x-1}{\frac{3}{2}-1} \\ & \frac{y-4}{\frac{1}{2}}=\frac{x-1}{\frac{1}{2}} \\ & x-y+3=0\end{aligned}$

Equation of median BE is

$\begin{array}{ll} & \frac{y-3}{5-3}=\frac{x-2}{1-2} \\ \therefore \quad & -1(y-3)=2(x-2) \\ \therefore & -y+3=2 x-4 \\ \therefore & 2 x+y=7\end{array}$

Equation of median CF is

$\begin{array}{ll} & \frac{y-6}{7}=\frac{x-1}{3} \\ & \frac{y-6}{2}-1 \\ \therefore \quad & \frac{y-6}{5}=\frac{x-1}{\frac{1}{2}} \\ & -\frac{x}{2} \\ \therefore \quad & y-6=-5(x-1) \\ \therefore \quad & 5 x+y-11=0\end{array}$

Equation of the line in two point form is $\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$

Equation of side AB is

$\begin{aligned} & \frac{y-4}{3-4}=\frac{x-1}{2-1} \\ & y-4=-1(x-1) \\ & y-4=-x+1 \\ & x+y=5\end{aligned}$

Equation of side BC is

$\frac{y-3}{6-3}=\frac{x-2}{1-2}$

-1(y – 3) = 3(x – 2) -y + 3 = 3x – 6 ∴ 3x + y = 9 Since both the points A and C have same x co-ordinates i.e. 1 the points A and C lie on a line parallel to Y-axis. ∴ The equation of side AC is x = 1.

By distance formula,

$\begin{aligned} & l(\mathrm{AB})=\sqrt{(4-0)^2+(3-0)^2}=\sqrt{25}=5 \\ & l(\mathrm{AC})=\sqrt{(4-2)^2+(3-3)^2}=\sqrt{4}=2\end{aligned}$

$\therefore \quad$ Point D divides $\mathrm{BC}$ internally in the ratio $5: 2$

$\therefore \quad \mathrm{D} \equiv\left(\frac{5(2)+2(0)}{5+2}, \frac{5(3)+2(0)}{5+2}\right) \equiv\left(\frac{10}{7}, \frac{15}{7}\right)$

$\therefore \quad$ Equation of $\mathrm{AD}$ is

$\frac{y-3}{\frac{15}{7}-3}=\frac{x-4}{\frac{10}{7}-4}$

$\therefore \quad \frac{y-3}{\frac{-6}{7}}=\frac{x-4}{\frac{-18}{7}}$

∴ 18 (y – 3) = 6 (x – 4)

∴ 3(y – 3) = x – 4

∴ 3y – 9 = x – 4

∴ x – 3y + 5 = 0

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is

given by u + kv = 0.

∴ (x +y – 2) + k(2x – 3y + 4) = 0 …(i)

But, x-intercept of line is 3.

∴ It passes through (3, 0).

Substituting x = 3 and y = 0 in (i), we get

(3 + 0 – 2) + k(6 – 0 + 4) = 0

∴ 1 + 10k = 0

$k=\frac{-1}{10}$

Substituting the value of $k$ in (i), we get $(x+y-2)+\left(\frac{-1}{10}\right)(2 x-3 y+4)=0$

∴ 10(x + y – 2) – (2x – 3y + 4) = 0

∴ 10x + 10y -20 — 2x + 3y-4 = 0

∴ 8x + 13y – 24 = 0, which is the equation of the required line.

Equation of the line passing through the point of intersection of lines u = 0 and v = 0 is given by u + kv = 0.

∴ (x + y – 2) + k(4x + 3y – 10) = 0 …(i)

∴ x + y – 2 + 4kx + 3ky – 10k = 0

∴ x + 4kx + y + 3ky – 2 – 10k = 0

∴ (1+ 4k)x + (1 + 3k)y – 2 – 10k = 0

But, this line is parallel to X-axis. ∴ Its slope = 0

$\begin{aligned} & \therefore \frac{-(1+4 k)}{1+3 k}=0 \\ & \therefore 1+4 k=0 \\ & \therefore k=\frac{-1}{4}\end{aligned}$

Substituting the value of k in (i), we get (x + y – 2) + (4x + 3y – 10) = 0 ∴ 4(x +y – 2) – (4x + 3y -10 ) = 0 ∴ 4x + 4y – 8 – 4x – 3y + 10 = 0 ∴ y + 2 = 0, which is the equation of the required line. [Note: Answer given in the textbook is 5y – 8= 0. However, as per our calculation it is y + 2 = 0.]

∴ x1 + y1 – 4 = 0

∴ y1 = 4 – x1 …(i)

Also, distance of P from the line 4x + 3y- 10 = 0 is 1

$\therefore \quad 1=\left|\frac{4 x_1+3 y_1-10}{\sqrt{4^2+3^2}}\right|$

$\therefore \quad 1=\left|\frac{4 x_1+3\left(4-x_1\right)-10}{\sqrt{25}}\right| \quad \ldots[$ From (i)]

$\therefore \quad 1=\left|\frac{4 x_1+12-3 x_1-10}{5}\right|$

∴ 5 = | x1 + 2 |

∴ x1 + 2 = ± 5

∴ x1 + 2 = 5 or x1 + 2 = – 5

∴ x1 = 3 or x1 = – 7 From (i),

when x1 = 3, y1 = 1

and when x1 = -7, y1 = 11

∴ The required points are (3, 1) and (-7, 11).

[Note: The question has been modified]

$\therefore \quad \frac{3 x}{\mathrm{p}}+\frac{4 y}{\mathrm{p}}=1$

$\therefore \quad \frac{x}{\mathrm{p}}+\frac{y}{\frac{\mathrm{p}}{3}}=1$

This equation is of the form $\frac{x}{a}+\frac{y}{b}=1$,

where $a=\frac{p}{3}$ and $b=\frac{p}{4}$

$\begin{aligned} & \therefore A(a, 0) \equiv\left(\frac{p}{3}, 0\right) \text { and } B \equiv(0, b)=\left(0, \frac{p}{4}\right) \\ & \therefore O A=\frac{p}{3} \text { and } O B=\frac{p}{4}\end{aligned}$

Given, A (∆OAB) = 24 sq. units

$\begin{aligned} & \therefore\left|\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\right|=24 \\ & \therefore\left|\frac{1}{2} \times \frac{\mathrm{p}}{3} \times \frac{\mathrm{p}}{4}\right|=24 \\ & \therefore \mathrm{p}^2=576 \\ & \therefore \mathrm{p}= \pm 24\end{aligned}$

$\therefore m_1=\frac{- \text { coefficient of } x}{\text { coefficient of } y}=\frac{-1}{-2}=\frac{1}{2}$

Let $m_2$ be the slope of the line $2 x+y+1=0$.

$\therefore \mathrm{m}_2=\frac{- \text { coefficient of } x}{\text { coefficient of } y}=\frac{-2}{1}=-2$

Since $m_1 \times m_1=\frac{1}{2} \times(-2)=-1$,

the given lines are perpendicular to each other. Consider,

x – 2y – 7 = 0 …(i)

2x + y + 1 =0 …(ii)

Multiplying equation (ii) by 2, we get

4x + 2y + 2 = 0 …(iii)

Adding equations (i) and (iii), we get

5x – 5 = 0

∴ x = 1

Substituting x = 1 in equation (ii), we get

2 + y + 1 = 0

∴ y = – 3

∴ The point of intersection of the given lines is (1,-3).

Slope of $A B=\frac{y_2-y_1}{x_1-x_1}=\frac{-2-(-1)}{4-3}=-1$

But, slope of a line = tan θ ∴ tan θ = – 1 = – tan 45° = tan (180° -45°) … [∵ tan (180° – θ) = -tan θ] = tan 135° ∴ θ = 135°

Let α be the acute angle that line AB makes with X-axis. Then, α + 0 = 180° α = 180°- 135° = 45° ∴ The acute angle between the X-axis and the line joining the points A and B is 45°.

$\therefore$ Slope of line $=\frac{y_2-y_1}{x_2-x_1}=\frac{3-0}{0-3}=-1$

But, slope of a line = tan θ ∴ tan θ = – 1

$\begin{aligned} & =-\tan \frac{\pi}{4} \\ & =\tan \left(\pi-\frac{\pi}{4}\right) \ldots[v \tan (\pi-\theta)=-\tan \theta] \\ & \tan \theta=\tan \frac{3 \pi}{4} \\ & \theta=\frac{3 \pi}{4}\end{aligned}$

The inclination of the line is $\frac{3 \pi}{4}$.

[Note: Answer given in the textbook is ' -1 However, as per our calculation it is $\frac{3 \pi}{4}$ ]

$\begin{aligned} & A P^2+P B^2=A B^2 P(x, y) \\ & \therefore\left[(x-1)^2+(y-6)^2\right]+\left[(x-3)^2+(y-5)^2\right]=(1-3)^2+(6-5)^2 \\ & \therefore x^2-2 x+1+y^2-12 y+36+x^2-6 x+9+y^2-10 y+25=4+1 \\ & \therefore 2 x^2+2 y^2-8 x-22 y+66=0 \\ & \therefore x^2+y^2-4 x-11 y+33=0\end{aligned}$

[Note: Answer given in the textbook is$3 x^2+4 y^2-4 x-11 y+33=0$

However, as per our calculation it is ' $x^2+y^2-4 x-11 y+33=0$