Question

# The point A divides the line segment joining the points (−5,1) and (3,5) in the ratio k:1. The coordinates of points B and C are (1,5) and (7,−2) respectively. If the area of △ABC is 2 square units, then the number of values of k is

A
1
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B
2
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C
3
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D
4
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Solution

## The correct option is B 2Given points are (−5,1) and (3,5) and the ratio is k:1, So, the point A=(3k−5k+1,5k+1k+1) Point B=(1,5), C=(7,−2) Now, the area of △ABC 2=12|x1(y2−y3)+x2(y3−y1)+x3(y1−y2)| ⇒2=12∣∣∣3k−5k+1(5+2)+1(−2−5k+1k+1)+7(5k+1k+1−5)∣∣∣⇒4=∣∣∣21k−35−5k−1+35k+7k+1−2−35∣∣∣⇒51k−29k+1−37=±4⇒14k−66k+1=±4⇒14k−66=±(4k+4)⇒14k−66=4k+4 or 14k−66=−4k−4∴k=7,319 Hence, the number of values of k is 2.

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