SATHEE: Straight Line and Pair of Straight Lines 1 Question 7

Straight Line and Pair of Straight Lines 1 Question 7

7. If a point $R (4, y, z)$ lies on the line segment joining the points $P (2, - 3, 4)$ and $Q (8, 0, 10)$ , then the distance of $R$ from the origin is

(2019 Main, 8 April II)

(a) $2 \sqrt{21}$

(b) $\sqrt{53}$

(d) 6

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Solution:

Given points are $P (2, - 3, 4), Q (8, 0, 10)$ and $R (4, y, z)$ . Now, equation of line passing through points $P$ and $Q$ is $\frac{x - 8}{6} = \frac{y - 0}{3} = \frac{z - 10}{6}$

[Since equation of a line passing through two points $A (x_{1}, y_{1}, z_{1})$ and $B (x_{2}, y_{2}, z_{2})$ is given by

$\begin{aligned} \frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \\ \Rightarrow \frac{x - 8}{2} = \frac{y}{1} = \frac{z - 10}{2} \end{aligned}$

$∵$ Points $P, Q$ and $R$ are collinear, so

$\begin{aligned} \frac{4 - 8}{2} & = \frac{y}{1} & = \frac{z - 10}{2} \\ \Rightarrow & - 2 & = y & = \frac{z - 10}{2} \\ \Rightarrow & y & = - 2 \\ and & = 6 \end{aligned}$

So, point $R$ is $(4, - 2, 6)$ , therefore the distance of point $R$ from origin is

$\begin{aligned} O R = \sqrt{16 + 4 + 36} \\ = \sqrt{56} = 2 \sqrt{14} \end{aligned}$

Straight Line and Pair of Straight Lines 1 Question 7

7. If a point $R (4, y, z)$ lies on the line segment joining the points $P (2, - 3, 4)$ and $Q (8, 0, 10)$ , then the distance of $R$ from the origin is

Solution:

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Straight Line and Pair of Straight Lines 1 Question 7

7. If a point R(4,y,z) lies on the line segment joining the points P(2,−3,4) and Q(8,0,10), then the distance of R from the origin 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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7. If a point $R (4, y, z)$ lies on the line segment joining the points $P (2, - 3, 4)$ and $Q (8, 0, 10)$ , then the distance of $R$ from the origin is