Question

Equation of a diameter of the circum circle of the triangle formed by the lines $3x+4y-7=0,3x-y+5=0and8x-6y+1=0$ is

• A. $3x-y-5=0$
• B. $3x+y+5=0$
• C. $3x-y+5=0$
• D. $3x+y-5=0$

Answer 1

The correct option is C $3x-y+5=0$

$L_1:3x+4y-7=0$ put $x=0$

$4y-7=0\therefore y=\frac{7}{4}A(0,\frac{7}{4})$

Put $y=0,x=\frac{7}{3}B(\frac{7}{3},0)$

$L_2:3x-y+5=0$

Put $x=0,y=5C(0,5)$

Put $y=0,x=\frac{-5}{3}D(\frac{-5}{3},0)$

$L_3:8x-6y+1=0$

Put $x=0,y=\frac{1}{6}E(0,\frac{1}{6})$

Put $y=0,x=\frac{-1}{8}F(\frac{-1}{8},0)$

Intersection point of $L_1$ and $L_3:$

$3x+4y-7=0\times 6$

$8x-6y+1=0\times 4$

$\Rightarrow 18x+24y-42=0\Rightarrow 32x+24y-4=0\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_50x-38=0$

$x=\frac{38}{50}$

$\therefore x=\frac{19}{25}$

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

$=\frac{7-3\times \frac{19}{25}}{4}$

$y=\frac{59}{50}$

$x=(\frac{19}{25},\frac{59}{50})$

Equation of the circle:

${(x+6)}^2+{(y-3)}^2=r^2$

$r=OA$

$=\sqrt{{(-6-4)}^2+{(3-7)}^2}$

$=\sqrt{{10}^2+4^2}$

$=\sqrt{100+16}$

$=\sqrt{116}$

$r=2\sqrt{29}$

Equation: ${(x+6)}^2+{(y-3)}^2=2\sqrt{29}$