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## GMAT - Problem Solving - Coordinate geometry

### The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ? (more)

The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ?

(A) k
(B) k/√2
(C) k/√3
(D) k/2
(E) k/3

Length from the origin to the point at which x axis touches the circle = radius of the circle = r (assume)

Length from the origin to the point at which y axis touches the circle = radius of the circle = r

Using pythagorus theorem

r2 + r2 = OC2 = k2

r = k/√2

B is correct

### Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6<= y<= 16. How many differ (more)

Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities -4 <= x <= 5 and 6<= y<= 16. How many different triangles with these properties could be constructed?

(A) 110
(B) 1,100
(C) 9,900
(D) 10,000
(E) 12,100

Number of ways of choosing coordinates of point P = number of ways of choosing a line with integral x coordinate between -4 and 5 * number of ways of choosing a line with integral y coordinate between 6 and 16 = 11*10 = 110

Since PR is parellel to x axis, y coordinate of R = y coordinate of P

Number of ways of choosing x coordinate of R = 9

Similarly number of ways of choosing y coordinate of Q = 10

Total number of ways of making a triangle = 110*9*10 = 9900

C is correct

### In the xy-coordinate plane, which of the following points must lie on the line kx + 3y = 6 for every possible value of k? (more)

In the xy-coordinate plane, which of the following points must lie on the line kx 3y = 6 for every possible value of k?

(A) (1,1)
(B) (0,2)
(C) (2,0)
(D) (3,6)
(E) (6,3)

The equation

kx + 3y = 6

can be written as the combination of two equations

kx = 0 -> x = 0

and 3y = 6 -> y = 2

(if we add both equations we get the equation of given line)

so the line must pass through (0, 2)

Alternatively

the equation kx + 3y = 6

will becoming independent of k only when x = 0, and since the point has to satisfy the equation for every value of k this has to be true.

so if x = 0

3y = 6 -> y = 2

thus the point will be (0, 2)

B. is correct

### In the rectangular coordinate system above, the area of triangular region PQR is (more)

In the rectangular coordinate system above, the area of triangular region PQR is

Area of a triangle is given by:

Mod $(\frac { 1 }{ 2 } ({ x }_{ 2 }{ y }_{ 3 }-{ y }_{ 2 }x_{ 3 }+{ x }_{ 3 }{ y }_{ 1 }-{ y }_{ 3 }{ x }_{ 1 }+{ x }_{ 1 }{ y }_{ 2 }-{ y }_{ 1 }{ x }_{ 2 }))$

= mod( 1/2(0*4 - 3*7 + 7*0 - 4*4 - 4*3 - 0*0)

= mod(1/2(-21 - 16 - 12)

=mod(1/2(-25)

= mod(-12.5)

= 12.5 units2

$r\quad =\sqrt { { (2-5) }^{ 2 }+{ (-3-0) }^{ 2 } } =\sqrt { 9+9 } =3\sqrt { 2 }$
$Area\quad =\quad \pi { r }^{ 2 }\quad =\quad \pi { (3\sqrt { 2 } ) }^{ 2 }=18\pi$