Let $PQR$ be a right angled isosceles triangle, right angled at $P (2,1)$. If the equation of the line $QR$ is $2x + y = 3$, then the equation representing the pair of lines PQ and PR is
- $ 3x^2 - 3 y^2 + 8 xy + 20 x + 10 y + 25 = 0 $
- $ 3x^2 - 3 y^2 + 8 xy - 20 x - 10 y + 25 = 0 $
- $ 3x^2 - 3 y^2 + 8 xy + 10 x + 15 y + 25 = 0 $
- $ 3x^2 - 3 y^2 - 8 xy + 10 x + 15 y + 25 = 0 $
The Correct Option is B
Solution and Explanation
Therefore, PS $ \perp$ QR and S is mid-point of hypotenuse,
$\therefore$ S is equidistant from P, Q, R.
$\therefore PS = QS = RS $
SInce, $ \angle P = 90^\circ$
and $ \angle Q = \angle R $
But $ \angle P + \angle Q + \angle R = 180^\circ $
$\therefore 90^\circ + \angle Q + \angle R = 180^\circ $
$\Rightarrow $ $ \angle Q = \angle R = 45^\circ$
Now, slope of QR is - 2 .
But $ QR \perp PS $
$\therefore $ Slope of PS is 1/2.
Let m be the slope of P
$\therefore \tan ( \pm 45^\circ) = \frac{ m - 1/ 2 }{ 1 - m \, ( - 1 / 2) }$
$\Rightarrow \pm 1 \frac{ 2m - 1}{ 2 + m} $
$\Rightarrow m = 3, - 1 / 3 $
$\therefore $ Equations of PQ and PR are $y - 1 = 3 (x - 2 )$ and $y - 1 = - \frac{1}{3} \, (x - 2)$
or 3 ( y - 1) + (x - 2) = 0
Therefore, joint equation of PQ and PR is
[ 3 (x - 2) - ( y - 1) ] [ (x - 2) + 3 ( y - 1) ] = 0
$\Rightarrow 3 ( x - 2 )^2 - 3 ( y - 1)^2 + 8 (x - 2) \, ( y - 1) = 0 $
$\Rightarrow 3x^2 - 3y^2 + 8 xy - 20 x - 10 y + 25 = 0 $
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Concepts Used:
Straight lines
A straight line is a line having the shortest distance between two points.
A straight line can be represented as an equation in various forms, as show in the image below:
The following are the many forms of the equation of the line that are presented in straight line-
1. Slope – Point Form
Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.
y – y0 = m (x – x0)
2. Two – Point Form
Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2) are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes
The slope of P2P = The slope of P1P2 , i.e.
\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)
Hence, the equation becomes:
y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)
3. Slope-Intercept Form
Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by
y – c =m( x - 0 )
As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if
y = m x +c