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II. Answer ALL the following questions: 03x03-099.Find the foot of the perpendicular of the point p(3,8)with respect to the line x + 3y - 7 = 0.10. Find the Angles between the lines V3x + y = 1 andx + 3y = 111. Reduce the equation v3x + y - 2 = 0 to the Normal fromXCOSA + ysina = P. |
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Answer» Answer: ANSWER Let line AB be x+3y=7 and point P be (3,8). Let Q(h,k) be the image of point P(3,8) in the line x+3y=7. Since line AB is a MIRROR, 1) Point P and Q are at equal distance from line AB, i.e., PR=QR, i.e., R is the mid-point of PQ. 2) Image is formed perpendicular to mirror i.e., line PQ is perpendicular to line AB. Since R is the MIDPOINT of PQ. Mid point of PQ joining (3,8) and (h,k) is ( 2 h+3
, 2 k+8
) Coordinate of point R = ( 2 h+3
, 2 k+8
) Since point R lies on the line AB. Therefore, ( 2 3+h
)+3( 2 8+k
)=7 h+3k=−13 ....(1) Also, PQ is perpendicular to AB. Therefore, SLOPE of PQ × Slope of AB=−1 Since, slope of AB= − 3 1
Therefore, slope of PQ= 3 Now, PQ is line joining P(3,8) and Q(h,k). Slope of PQ = 3= h−3 k−8
3h−k=1 ........(2) Solving equation 1 and 2, we get, h=−1 and k=−4 Hence, image is Q(−1,−4). |
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