GMAT Club Around The World!!! In the coordinate plane above, if line

In the coordinate plane above, if line 1 || line 2 || x-axis and point a on line 2 is exactly 13 units away from point b on line 1, what is the x-coordinate of point b ?

A. 19
B. 17
C. 15
D. 12
E. 8


x coordinate b = 7 + underroot (13^2- 5^2) = 7+ 12 = 19

ans = A)19

Point a is 13 units away from point b, so we get a right triangle with hypotenuse = 13.
The perpendicular distance between line a and line b is 8-3 = 5.
So, the horizontal distance must be 12 units (Pythagorean triplets).
Point b must be 12 + 7 = 19 units away from origin. Therefore, its x- coordinate should also be 19.

Answer A.

Can't wait to read the explanations here. I have no precise method...

I guess it's C.

Ans A

Equation of Line 1 Y = 8 and Line 2 is Y = 3

The Y co-ordinate of each of the lines is 8 for L1 and 3 for L2

So Point a = (7,3) and b = (X, 8)

Given the distance between a and b is 13

Therefore Sqrt((X-7)^2 * (8-3)^2) = 13

>> X^2 - 14X - 95 = 0

>> X = 19 or -5

In PS questions, we can trust the figure given unless otherwise is specified. Point b is in Q1, so X cannot be negative.

The answer is A

The vertical distance between a & b is 5 unit and the hypotenuse is 13, therefore the horizontal distance is √(169-25) = 12

So, x-coordinate of point b is 7+12 = 19

IMO A

distance between a and b = 13
triangle ABC is a right angle triangle right angled at B
AC = 13
BC = 8 - 3 = 5
AB = \(√13^2-5^2\) = 12
x coordinate of b = x coordinate of B = x coordinate of a + AB = 7 + 12 = 19

Given, distance between a and b= 13

a= (7,3), so, 13^2 = (x-7)^2 + (8-3)^2 or, (x-7)^2 = 169-25= 144, x-7= 12, (consider in positive x)

so, x= 19. So, I think A.

The straight line between a and b forms the hypotenuse of a right angled triangle formed by perpendicular line from b to line l1, whose length is given as 13 units.
The vertical distance between l1 and l2 is difference in y co-ordinates i.e. 8-3 = 5.
The horizontal distance between a and b will be \(\sqrt{13^2 - 5^2} = \sqrt{12^2}\)
Thus the x co-ordinate of b will be x co-ordinate of a (7) + 12 = 19.
Answer: A

IMO:A = 19
Given, distance btw A & B is 13units i.e. the shortest distance.
Hence, according to Pythagoras theorem => 13^2 = 5^2+(horizontal distance btw A & B)^2
 horizontal distance btw A & B = 12
 therefore, x unit of B = 12+7 = 19

Correct Answer A

Coordinate of a( 7, 3) and b (x+7, 8)
it will form the right angle triangle
D = 13
13^2 = 5^2 +x^2
x = 12

X coordinate of b = 12+7 = 19

ImoA

If a point is 13 unit away
Means
13^2=5^2-x^2
X=12

B cordinate of x=12+7=19

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From the figure, we know that y coordinate of line 1 is 8 and it will be 8 for all points on the line as its a horizontal line parallel to the x-axis
Same way y coordinate of line 2 is 3 and it will be 3 for all points on the line as its a horizontal line parallel to the x-axis
From the figure, we also know that the x coordinate for a is 7

The distance between a and b is 13

So using the distance between 2 points formula we can write
13 = √(x-7)^2 +(8-3)^2
13^2 = (x-7)^2 + 5^2
169 = x^2 +49 -14x +25
169 = x^2 -14x + 74
x^2 -14x -95 = 0
(x-19) (x+5) = 0
x = 19 , -5

Since point b is in the first quadrant so the value of x will be positive which is 19

Hence the answer will be A

Let the x-coordinate of point b be x.

Co-ordinates of point A: (7,3) and point B: (x,8).

From the diagram and information provided in the question statement:

(x-7)^2 + (8-3)^2 = (13)^2

-> (x-7)^2 = (12)^2
-> x-7 = 12 / -12
-> x = 19 /-5

Since, point B is in the first quadrant, x>0.
thus, x = 19.

IMO, option A is the answer.

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