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The graph of a line in the 
xy-plane has a slope of 1 and contains the point 
(3,0). The graph of a second line has a slope of 
-(1)/(4) and contains the point 
(-7,0). If the two lines intersect at the point 
(a,b), what is the value of 
a+b ?
Choose 1 answer:
(A) -8
(B) -5
(C) -2
(D) -1

The graph of a line in the xy x y -plane has a slope of 11 and contains the point (3,0) (3,0) . The graph of a second line has a slope of 14 -\frac{1}{4} and contains the point (7,0) (-7,0) . If the two lines intersect at the point (a,b) (a, b) , what is the value of a+b a+b ?\newlineChoose 11 answer:\newline(A) 8-8\newline(B) 5-5\newline(C) 2-2\newline(D) 1-1

Full solution

Q. The graph of a line in the xy x y -plane has a slope of 11 and contains the point (3,0) (3,0) . The graph of a second line has a slope of 14 -\frac{1}{4} and contains the point (7,0) (-7,0) . If the two lines intersect at the point (a,b) (a, b) , what is the value of a+b a+b ?\newlineChoose 11 answer:\newline(A) 8-8\newline(B) 5-5\newline(C) 2-2\newline(D) 1-1
  1. Find equation of first line: First, let's find the equation of the first line with a slope of 11 that passes through the point (3,0)(3,0).\newlineThe slope-intercept form of a line is y=mx+by = mx + b, where mm is the slope and bb is the y-intercept.\newlineSince the slope (mm) is 11 and the line passes through (3,0)(3,0), we can substitute these values into the equation to find bb.\newliney=mx+by = mx + b\newline0=(1)(3)+b0 = (1)(3) + b\newline0=3+b0 = 3 + b\newlineb=3b = -3 \newlineNow, substitute the value of mm and bb in y=mx+by = mx + b.\newlineSo, the equation of the first line is y=x3y = x - 3.
  2. Find equation of second line: Next, let's find the equation of the second line with a slope of 14-\frac{1}{4} that passes through the point (7,0)(-7,0). Again, using the slope-intercept form y=mx+by = mx + b, we substitute the slope and the point into the equation to find bb. \newliney=mx+by = mx + b \newline0=14(7)+b0 = -\frac{1}{4}(-7) + b \newline0=74+b0 = \frac{7}{4} + b \newlineb=74b = -\frac{7}{4} \newlineNow, substitute the value of mm and bb in y=mx+by = mx + b. \newlineSo, the equation of the second line is y=14x74y = -\frac{1}{4}x - \frac{7}{4}.
  3. Find intersection point: Now, to find the intersection point (a,b)(a,b) of the two lines, we set their equations equal to each other since at the point of intersection, their yy-values will be the same.x3=(14)x74x - 3 = -\left(\frac{1}{4}\right)x - \frac{7}{4}To solve for xx, we combine like terms.x+(14)x=74+3x + \left(\frac{1}{4}\right)x = -\frac{7}{4} + 3(54)x=54\left(\frac{5}{4}\right)x = \frac{5}{4}x=54×45x = \frac{5}{4} \times \frac{4}{5}x=1x = 1
  4. Find coordinates of intersection point: Having found the xx-coordinate of the intersection point, we now substitute x=1x = 1 into either of the line equations to find the yy-coordinate. We'll use the first line's equation for simplicity.y=x3y = x - 3y=13y = 1 - 3y=2y = -2So, the intersection point (a,b)(a,b) is (1,2)(1,-2).
  5. Find value of a+ba + b: Finally, we find the value of a+ba + b.\newlinea+b=1+(2)a + b = 1 + (-2)\newlinea+b=1a + b = -1

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