Geometry problems

Q1 : 5.14 [36/7]

Q2 : (-1,-2)

I 've seen them somewhere. Is the source Manhattan GMAT ?

Pls confirm the answers...

Yes Gughan,

They are from manhattan and your answers are correct.

can someone explain the solution of this problem

'manpreet singh wrote:Q.1 Find x in the attached diagram?

Any side of a triangle can be considered the base.
Each base has a corresponding height.
A = 1/2(bh).
Since the area must be the same no matter which base and height are used, bh must always yield the same product.
In the triangle above:
If the side of 12 is considered the base, the corresponding height is 3.
If the side of 7 is considered the base, the corresponding height is x.
Thus:
3*12 = 7x
x = 36/7.

Q.2 The line represented by the equation y = -2x + 6 is the perpendicular bisector of the line segment AB. If A has the coordinates (7, 2), what are the coordinates for B?

The most efficient approach is to DRAW and to use the answer choices, which an actual GMAT problem would provide.
Check my posts here:

https://www.beatthegmat.com/grockit-toug ... 74452.html

https://www.beatthegmat.com/og-12-t77036.html

Q.2 : The line represented by the equation y = -2x + 6 is the perpendicular bisector of the line segment AB. If A has the coordinates (7, 2), what are the coordinates for B?

Solution :

The line represented by the equation y = -2x + 6 is the perpendicular bisector of the line segment AB. So, The line segment AB and the line y = -2x + 6 should have a common point. That common point is the point of intersection of the line y = -2x + 6 and the line segment AB. To find that common point, we have to solve the 2 line equations. We have one line equation -> y = -2x + 6. To find the other we follow these steps :

1.Slope : Since, these two lines are perpendicular, the product of their slopes = -1. The line, y = -2x + 6, has a slope = -2 [y=mx+c]. So, the perpendicular line should have a slope = (1/2). Hence the product is -1.

Line Equ 2 = y=(1/2)x+c

2.y-intercept(c) : Substitute the point A(7, 2) in y=(1/2)x+c.

2=(1/2)7+c
2=(7/2)+c
2-(7/2)=c
(4-7)/2=c
c=(-3/2)

Line equ 2 = y=(1/2)x-(3/2)

Now, we have the 2 equations. We have to solve them.

y=-2x+6 ------ 1
y=(1/2)x-(3/2) --------- 2

-2x+6=(1/2)x-(3/2)
-4x+12=x-3
15=5x
x=3

Sub x=3 in 1

y=-2(3)+6
y=-6+6
y=0

We 've now got the midpoint of AB [The common point] = (3,0)

To find B. Since they are equally spaced. We can follow this

A(7,2) B(x,y) common point(3,0)

(7+x)/2 = 3
7+x=6
x=-1

Similarly,

(2+y)/2=0
2+y=0
y=-2

Hence, the point B is (-1,-2).

Hope this clears the doubt