Straight Lines Question - None

Question

Slope of a line passing through P(2, 3) and intersecting the line, x + y = 7 at a distance of 4 units from P, is

Accepted Answer

Let the slope of line is m, which is passing through P(2, 3)

Since, the distance of a point (x₁, y₁) from the line ax + by + c = 0 is d = $\left|\frac{a x_{1}+b y_{1}+c}{\sqrt{a^{2}+b^{2}}} ight|$
$ herefore$ The distance of a point P(2, 3) from the line x + y - 7 = 0, is d = $\frac{| 2+3-7|}{\sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2}$
Now, is $ riangle$PQR,
QR =$\sqrt{16-d^{2}}=\sqrt{16-2}=\sqrt{14}$
$ herefore an heta=\frac{d}{Q R}=\frac{\sqrt{2}}{\sqrt{14}}=\frac{1}{\sqrt{7}}=\left|\frac{m+1}{1-m} ight|$ $\left[\because an heta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}} ight| ight]$
$\Rightarrow \frac{m+1}{1-m}=\pm \frac{1}{\sqrt{7}}$
$\Rightarrow \frac{m+1}{1-m}=\frac{1}{\sqrt{7}} ext { or } \frac{m+1}{1-m}=-\frac{1}{\sqrt{7}}$
$\Rightarrow m=\frac{1-\sqrt{7}}{1+\sqrt{7}} ext { or } m=\frac{-1-\sqrt{7}}{\sqrt{7}-1}$

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