Line $p$ lies on a coordinate plane. What is the area of the region bounded by the $x$-axis, the $y$-axis and line

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

Line $p$ lies on a coordinate plane. What is the area of the region bounded by the $x$-axis, the $y$-axis and line

by Vincen » Fri Jun 19, 2020 2:39 am

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Line $p$ lies on a coordinate plane. What is the area of the region bounded by the $x$-axis, the $y$-axis and line $p?$

(1) The slope of line $p$ is $-\dfrac53.$

(2) The $y$-intercept of line $p$ is $10.$

[spoiler]OA=C[/spoiler]

Source: Veritas Prep

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Source: Beat The GMAT — Data Sufficiency | Fri Jun 19, 2020 2:39 am

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pm; Followed by:5 members

Re: Line $p$ lies on a coordinate plane. What is the area of the region bounded by the $x$-axis, the $y$-axis and

by deloitte247 » Sat Jun 20, 2020 5:05 am

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Target question => What is the area of the region bounded by x-axis, y-axis, and line p?
The region described in the question stem will form a right angle triangle
Area of right-angle triangle = 1/2 * base * height
Where base = x-intercept and height = y-intercept

$$Statement\ 1\ =>\ The\ slope\ of\ the\ line\ is\ -\frac{5}{3}$$
To find x-intercept from equation of line, substitute y = 0 and express in terms of x
$$y=mx+c\ where\ y=0\ and\ m\left(slope\right)=-\frac{5}{3}$$
$$0=-\frac{5}{3}x+c$$
$$+\frac{5}{3}x=c$$
$$x=c\div\frac{5}{3}\ =c\cdot\frac{3}{5}=>\frac{3c}{5}$$
But x-intercept is unknown hence statement 2 is NOT SUFFICIENT

Combining both statements together =>
y - intercept = 10
x - intercept = 3c/5 where c = y-intercept = 10
x-intercept = 3 * 10/5 = 6

Area = 1/2 * 6 * 10 = 1/2 * 60 = 30
Both statements together combined together ARE SUFFICIENT
Answer = C