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put this solution on YOUR website!1. Write in slope intercept the equation of theline passing through the two points.Show that the line is perpindicular to the given line.
(-2,-2), (1,3); y= 3x-1
EQN.OF LINE JOINING (X1,Y1) AND (X2,Y2) IS GIVEN BY
Y-Y1=(X-X1)*(Y2-Y1)/(X2-X1), WHERE (Y2-Y1)/(X2-X1) IS THE SLOPE OF THE LINE AND Y1-X1*(Y2-Y1)/(X2-X1)IS THE INTERCEPT.
HENCE EQN.OF LINE IS
Y+2=(X+2)*(3+2)/(1+2)=X(5/3)+2*5/3
Y=X(5/3)+10/3-2=X(5/3)+4/3.....HENCE SLOPE IS 5/3 AND INTERCEPT IS 4/3
SLOPE OF GIVEN LINE Y=3X-1 IS 3 ,,,FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HERE THE PRODUCT IS 3*5/3=5..HENCE THEY ARE NOT PERPENDICULAR TO EACH OTHER..CHECK BACK YOUR NUMBERS...COPY THE PROBLEM PROPERLY.AS GIVEN THE LINES ARE NOT PERPENDICULAR.
2. write in slope-intercept form the equation of the line passing through the given point and perpindicular to the given line.
(-4,-7), y=-4x-7
SLOPE OF Y=-4X-7 IS -4
AS GIVEN ABOVE FOR 2 LINES TO BE PERPENDICULAR,THE PRODUCT OF THEIR SLOPES SHOULD BE -1.HENCE THE SLOPE OF THE REQUIRED LINE IS -1/-4=1/4
EQN.OF REQD. LINE IS
Y+7=(1/4)(X+4)=X/4 + 1
Y= X/4 - 6
3.Write in standard form an equation of the line that passes through the two points. Use interger coefficients.
(-1,0) and (3,10)
SAME AS THE FIRST PROBLEM ABOVE...
