Math Forum :: View topic – More about STRAIGHT LINES………

1. From x+y=2 and 3x^2+y^2=6, the intersection points A and B can be found. M can then be found. Slope of the required line can be found from that of AB. Hence, by point-slope form we can find the required line. 2. Point P can be found from A and B in terms of r. Since slope of AB and OP are known and their product =-1, r can then be found. 3. Find the intersection point, P(a,b) between AB and x-y+1=0 first, by a=[r(-1)+3] / (r+1), the ratio can be found. 4. Find the 3 intersection points A, B and C between L1, L2 and L3 first. Then the area of ABC can be expressed in terms of c. By equating it to 4, c can be found. 5. a) Let D be the foot. Since BC is perpendicular to AD and slope of AD can be found, slope of BC can be found. With D lies on BC, by point-slope form to find BC. b)

Since B lies on BC, the value of a can be found. Also, the y-coorindate of C, relates with its x-coordinate by the line BC. Let C=(b,f(b)). With slope of BA slope of AC = -1, b and hence coorindates of C can be found.