Pls solve this equation?

Question

the stright line L1 has equation 2x + y - 14 = 0 and crosse the x-axis at the point A

(a) Find the coordinates of A

the stright line L2 is parallel to L1 and passes through the point B(-6,6)

(b)find an equation for L2 in the form of y= mx +c

The line L2 crosses the x-axis at the point C

(c)find the coordinates of C

The point D lies on L1 and is such that CD is perpendicular to L1

(d)show that d has coodinates (5,4)

(e)find the area of triangle ACD

Answer 1

i know the answers for (a), (b) and (c)

(a) An equation passing through the x-axis will always have the co-ordinates as (a,0) => y=0

substituting y=0 in L1

=> 2x-14=0

=> x=7

therefore the co-ordinates of A is (7,0)

(b) the line L2 is parallel to L1 => the slope of L2 = the slope of L1

slope of L1 = (-2)/1 = (-2) = slope of L2

the line passes through the point (6,-6)

the slope point formula is (y2-y1) = m(x2-x1) ------- [ m is the slope and y2 means y subscript 2]

=> y-6 = -2(x+6)

=> y-6 = -12-2x

=> y = -2x-6 which is of the form y= mx+c ------ [ m = -2 and c = -6 ]

(c) since the line L2 passes through point C we can substitute coordinates of C in the line eqn

since C is a point on the x-axis, the value of y is 0

substituting y=0 in the eqn L2

=> 0 = -2x - 6

=> x = -8

the coordinates of C is ( -8,0 )

Im not able to get the other two answers.. sorry!