Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA5 Marks
Question
Show that the points (3, 4), (-5, 16) and (5, 1) are collinear.
✓
Answer
Here, let A = (3, 4)
B = (-5, 16)
C = (5, 1)
$\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A
$=\big(-5\hat{\text{i}}+16\hat{\text{j}}\big)-\big(3\hat{\text{i}}+4\hat{\text{j}}\big)$
$=-5\hat{\text{i}}+16\hat{\text{j}}-3\hat{\text{i}}-4\hat{\text{j}}$
$\overrightarrow{\text{AB}}=-8\hat{\text{i}}+12\hat{\text{j}}$
$\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B
$=\big(5\hat{\text{i}}+\hat{\text{j}}\big)-\big(-5\hat{\text{i}}+16\hat{\text{j}}\big)$
$=5\hat{\text{i}}+\hat{\text{j}}+5\hat{\text{i}}-16\hat{\text{j}}$
$\overrightarrow{\text{BC}}=10\hat{\text{i}}-15\hat{\text{j}} $
So, $4\Big(\overrightarrow{\text{AB}}\Big)=-5\Big(\overrightarrow{\text{BC}}\Big)$
$\overrightarrow{\text{AB}}$ is parallel to $\overrightarrow{\text{BC}}$ but B is a common point.
Hence, A, B, C are collinear.
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