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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

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle? (more)

In the coordinate plane, a circle has center (2, -3) and passes through the point (5, 0). What is the area of the circle?

Radius of the circle = distance between the center and the point through which the circle passes

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