Straight Lines
Grade 11

Question:

<p>If a point R(4, y, z) lies on the line segment joining the points P{2, -3, 4) and Q(8,&nbsp;0, 10), then the distance of R from the origin is</p>
<p style="display:inline">6</p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{21}\)</span></p>
<p style="display:inline"><span class="math-tex">\(2 \sqrt{14}\)</span></p>
<p style="display:inline"><span class="math-tex">\(\sqrt{53}\)</span></p>

Step-by-Step Solution

Key Concept: Use the symmetric form of a line passing through two points to find the unknown coordinates of a collinear point by equating the ratios.
<p>Given points are P(2, -3,4), Q(8,0,10) and R(4, y, z).<br /> Now, equation of line passing through points P and Q is<br /> <span class="math-tex">\(\frac{x-8}{6}=\frac{y-0}{3}=\frac{z-10}{6}\)</span>&nbsp;&nbsp;[Since equation of a line passing through two points A(x<sub>1</sub>, y<sub>1</sub>, z<sub>1</sub>) and B(x<sub>2</sub>, y<sub>2</sub>, z<sub>2</sub>) is given by&nbsp;<span class="math-tex">\(\left.\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\right]\)</span><br /> <span class="math-tex">\(\Rightarrow \quad \frac{x-8}{2}=\frac{y}{1}=\frac{z-10}{2}\)</span>&nbsp;...(i)<br /> <span class="math-tex">\(\because\)</span>&nbsp;Points P, Q and P are collinear, so<br /> <span class="math-tex">\(\frac{4-8}{2}=\frac{y}{1}=\frac{z-10}{2}\)</span><br /> <span class="math-tex">\(\Rightarrow\)</span>&nbsp;-2 = y =&nbsp;<span class="math-tex">\(\frac{z-10}{2}\)</span><br /> <span class="math-tex">\(\Rightarrow\)</span>&nbsp;y = -2 and z = 6<br /> So, point R is (4, - 2, 6), therefore the distance of point R from origin is<br /> OR =&nbsp;<span class="math-tex">\(\sqrt{16+4+36}\)</span><br /> <span class="math-tex">\(=\sqrt{56}=2 \sqrt{14}\)</span></p>
Correct Answer: C

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