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Bunuel
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Line L passes through the origin and is perpendicular to line M, which 11 Feb 2023, 23:20
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85% (hard)

Question Stats:

56% (02:43) correct 44% (02:41) wrong based on 146 sessions

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Line L passes through the origin and is perpendicular to line M, which is defined by the equation y = 6 - 2x. What is the area of the triangle formed by the y-axis , L, and M ?

A. 1.8
B. 3.6
C. 7.2
D. 9
E. 12
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C
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Line L passes through the origin and is perpendicular to line M, which 12 Feb 2023, 00:36
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Bunuel
Line L passes through the origin and is perpendicular to line M, which is defined by the equation y = 6 - 2x. What is the area of the triangle formed by the y-axis , L, and M ?

A. 1.8
B. 3.6
C. 7.2
D. 9
E. 12
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y intercept of line M = (0 , 6) ⇒ This point is indicated by P

y = -2x + 6

The slope of line M = -2

Slope of a line perpendicular to M = \(\frac{1}{2}\)

As L is a straight line, it can be represented by the equation

\(y - y_1 = m (x - x_1)\)

we know the slope of line L = \(\frac{1}{2}\)

As the line passes through the origin \(x_1 = y_1 = 0\)

Substituting the values we get

\(y = \frac{x}{2}\)

Let's find the point of intersection of line M and line L, we can indicate the point of intersection by Q

y = -2x + 6

x/2 = -2x + 6

Solving further we get, x = \(\frac{12}{5}\). Hence the x coordinates of Q = \(\frac{12}{5}\)

Area of \(\triangle POQ\) =

\(\frac{1}{2}\) * base * height

\(\frac{1}{2}\) * PO *height

\(\frac{1}{2}\) * PO * x-coordinate of Q

\(\frac{1}{2}\) * 6 * \(\frac{12}{5}\)

= 36/5

= 7.2

Option C
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Line L passes through the origin and is perpendicular to line M, which 14 Jul 2025, 05:13
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till finding x-coordinate of Q = 12/5, i was following your solution, but how did you calculate the area like this? Area = 1/2 * base * height, base is OQ and height is PQ, dont we need to use distance formula here for this? Why are we bringing length of PO which is a hypotenuse of triangle POQ in the calculation of area. I'm totally confused here. Please someone help Bunuel chetan2u GMATNinja
gmatophobia
Bunuel
Line L passes through the origin and is perpendicular to line M, which is defined by the equation y = 6 - 2x. What is the area of the triangle formed by the y-axis , L, and M ?

A. 1.8
B. 3.6
C. 7.2
D. 9
E. 12

y intercept of line M = (0 , 6) ⇒ This point is indicated by P

y = -2x + 6

The slope of line M = -2

Slope of a line perpendicular to M = \(\frac{1}{2}\)

As L is a straight line, it can be represented by the equation

\(y - y_1 = m (x - x_1)\)

we know the slope of line L = \(\frac{1}{2}\)

As the line passes through the origin \(x_1 = y_1 = 0\)

Substituting the values we get

\(y = \frac{x}{2}\)

Let's find the point of intersection of line M and line L, we can indicate the point of intersection by Q

y = -2x + 6

x/2 = -2x + 6

Solving further we get, x = \(\frac{12}{5}\). Hence the x coordinates of Q = \(\frac{12}{5}\)

Area of \(\triangle POQ\) =

\(\frac{1}{2}\) * base * height

\(\frac{1}{2}\) * PO *height

\(\frac{1}{2}\) * PO * x-coordinate of Q

\(\frac{1}{2}\) * 6 * \(\frac{12}{5}\)

= 36/5

= 7.2

Option C
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svsivakrishna
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Line L passes through the origin and is perpendicular to line M, which 15 Jul 2025, 11:33
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Hi,

The height can be considered as 12/5 (this is the value of x coordinate and a line drawn will be parallel to x axis and perpendicular to y axis which acts as height for the triangle) and the base can be considered as OP= 6, therefore area = 1/2 * 6 * 12/5 =7.2
dejavu3363
till finding x-coordinate of Q = 12/5, i was following your solution, but how did you calculate the area like this? Area = 1/2 * base * height, base is OQ and height is PQ, dont we need to use distance formula here for this? Why are we bringing length of PO which is a hypotenuse of triangle POQ in the calculation of area. I'm totally confused here. Please someone help Bunuel chetan2u GMATNinja
gmatophobia
Bunuel
Line L passes through the origin and is perpendicular to line M, which is defined by the equation y = 6 - 2x. What is the area of the triangle formed by the y-axis , L, and M ?

A. 1.8
B. 3.6
C. 7.2
D. 9
E. 12

y intercept of line M = (0 , 6) ⇒ This point is indicated by P

y = -2x + 6

The slope of line M = -2

Slope of a line perpendicular to M = \(\frac{1}{2}\)

As L is a straight line, it can be represented by the equation

\(y - y_1 = m (x - x_1)\)

we know the slope of line L = \(\frac{1}{2}\)

As the line passes through the origin \(x_1 = y_1 = 0\)

Substituting the values we get

\(y = \frac{x}{2}\)

Let's find the point of intersection of line M and line L, we can indicate the point of intersection by Q

y = -2x + 6

x/2 = -2x + 6

Solving further we get, x = \(\frac{12}{5}\). Hence the x coordinates of Q = \(\frac{12}{5}\)

Area of \(\triangle POQ\) =

\(\frac{1}{2}\) * base * height

\(\frac{1}{2}\) * PO *height

\(\frac{1}{2}\) * PO * x-coordinate of Q

\(\frac{1}{2}\) * 6 * \(\frac{12}{5}\)

= 36/5

= 7.2

Option C
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