Let the lines $(y - 2) = m_{1} (x - 5)$ and $(y + 4) = m_{2} (x - 3)$ intersect at right angles at $P$ (where $m_{1}$ and $m_{2}$ are parameters). If locus of $P$ is, $x^{2} + y^{2} + g x + f y + 7 = 0$ , then $| f + g | = ?$

Open in App

Solution

$(y - 2) = m_{1} (x - 5)$ ......... $(1)$ $m_{1} m_{2} = - 1$
$(y - 4) = m_{2} (x - 3)$ ........... $(2)$
multiply $(1)$ & $(2)$
$(y - 2) (y - 4) = (x - 5) (x - 3)$
$y^{2} - 6 y + 8 = [x^{2} + 15 - 8 x]$
$x^{2} + y^{2} - 8 x - 6 x + 7 = 0$

Suggest Corrections

Similar questions

y - 1 = m_{1} (x - 3)

and

y - 3 = m_{2} (x - 1)

are two family of straight lines, at right angles to each other. The locus of their point of intersection is

Q. If the lines (y-b)=

m_{1}

(x+a) and (y-b)=

m_{2}

(x+a) are the tangents of

y^{2} = 4 a x

then

Q. If

m_{1}

and

m_{2}

are the roots of the equation

x^{2} + (\sqrt{3} + 2) x + (\sqrt{3} - 1) = 0

, then the area of the triangle formed by the lines

y = m_{1} x, y = m_{2} x

and

y = 2

is :

Q. If

y + b = m_{1} (x + a), y + b = m_{2} (x + a)

are two tangents to the parabola

y^{2} = 4 a x,

then

Show that the are of the triangle formed by the lines $y = m_{1} x, y = m_{2}$ x and y = c is equal to $\frac{c^{2}}{4} (\sqrt{33} + \sqrt{11})$ , where $m_{1}, m_{2}$ are the roots of the equation $x^{2} + (\sqrt{3} + 2) x + \sqrt{3} - 1 = 0.$

Join BYJU'S Learning Program

Let the lines (y−2)=m1(x−5) and (y+4)=m2(x−3) intersect at right angles at P (where m1 and m2 are parameters). If locus of P is, x2+y2+gx+fy+7=0, then |f+g|=?

(y−2)=m1(x−5) ......... (1) m1m2=−1(y−4)=m2(x−3) ........... (2)multiply (1) & (2)(y−2)(y−4)=(x−5)(x−3)y2−6y+8=[x2+15−8x]x2+y2−8x−6x+7=0

Let the lines $(y - 2) = m_{1} (x - 5)$ and $(y + 4) = m_{2} (x - 3)$ intersect at right angles at $P$ (where $m_{1}$ and $m_{2}$ are parameters). If locus of $P$ is, $x^{2} + y^{2} + g x + f y + 7 = 0$ , then $| f + g | = ?$

$(y - 2) = m_{1} (x - 5)$ ......... $(1)$ $m_{1} m_{2} = - 1$
$(y - 4) = m_{2} (x - 3)$ ........... $(2)$
multiply $(1)$ & $(2)$
$(y - 2) (y - 4) = (x - 5) (x - 3)$
$y^{2} - 6 y + 8 = [x^{2} + 15 - 8 x]$
$x^{2} + y^{2} - 8 x - 6 x + 7 = 0$