SATHEE: Straight Line and Pair of Straight Lines 1 Question 6

Straight Line and Pair of Straight Lines 1 Question 6

6. 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

(2019 Main, 9 April I)

(a) $\frac{1 - \sqrt{5}}{1 + \sqrt{5}}$

(b) $\frac{\sqrt{7} - 1}{\sqrt{7} + 1}$

(d) $\frac{\sqrt{5} - 1}{\sqrt{5} + 1}$

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Solution:

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

Since, the distance of a point $(x_{1}, y_{1})$ from the line $a x + b y + c = 0$ is $d = | \frac{a x_{1} + b y_{1} + c}{\sqrt{a^{2} + b^{2}}} |$ .

$∴$ 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, in $△ P R Q$ ,

$Q R = \sqrt{16 - d^{2}} = \sqrt{16 - 2} = \sqrt{14}$

$\begin{array}{llrl} ∴ & \tan θ = \frac{d}{Q R} & = \frac{\sqrt{2}}{\sqrt{14}} = \frac{1}{\sqrt{7}} = | \frac{m + 1}{1 - m} | \\ \Rightarrow & \frac{m + 1}{1 - m} & = \pm \frac{1}{\sqrt{7}} \\ \Rightarrow & \frac{m + 1}{1 - m} & = \frac{1}{\sqrt{7}} or \frac{m + 1}{1 - m} = - \frac{1}{\sqrt{7}} \\ \Rightarrow & m & = \frac{1 - \sqrt{7}}{1 + \sqrt{7}} or m = \frac{- 1 - \sqrt{7}}{\sqrt{7} - 1} \end{array}$

Straight Line and Pair of Straight Lines 1 Question 6

6. 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

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Straight Line and Pair of Straight Lines 1 Question 6

6. 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

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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6. 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