Question

If the slope of a line passing through the point $A(3,2)$ be $\frac{3}{4}$, then the points on the line which are $5$ units away from $A$ are

• A. $(5,5),(-1,-1)$
• B. $(7,5),(-1,-1)$
• C. $(5,7),(-1,-1)$
• D. $(7,5),(1,1)$

Answer 1

The correct option is B $(7,5),(-1,-1)$

Equation of line with given slope $\frac{3}{4}$ and from point $A(3,2)$

$y-2=\frac{3}{4}(x-3)$

$4y-8=3x-9$

$3x=1+4y$------(1)

Let Point $B(x_1,y_1)$

The distance between A and B is 5 Hence by distance formula

$(x_1-3{)}^2+(y_1-2{)}^2=25$

$x_1^2+9-6x_1+y_1^2+4-4y_1=25$

$x_1^2+y_1^2-6x_1-4y_1-12=$

Point B lies on line $3x=1+4y$

Hence $x_1=\frac{4y_1+1}{3}$

${\left(\frac{4y_1+1}{3}\right)}^2+y_1^2-6\left(\frac{4y_1+1}{3}\right)-4y_1-12=0$

$\frac{16y_1^2+1+8y_1}{9}+y_1^2-8y_1-2-4y_1-12=0$

$16y_1^2+1+8y_1+9y_1^2-72y_1-18-36y_1-108=0$

$25y_1^2-100y_1-125=0$

$y_1^2-4y_1-5=0$

$y_1^2-5y_1+y_1-5=0$

$(y_1-5)(y_1+1)=0$

$\Rightarrow y_1=-1,5$

From equation (1)

$3x_1=1+4y_1$

$3x_1=1-4$ and $3x_1=1+20$

$x_1=-1$ and $x_1=7$

Point B$(7,5)$ and $(-1,-1)$