Slope and y-intercept

The line is \(y=-6x+c\) and it passes through \((5,-18)\). Substituting \(x=5\) and \(y=-18\) gives \(-18=-6(5)+c=-30+c\). Hence, \(c=12\). In the form \(y=mx+c\), \(c\) is the \(y\)-intercept, so the correct answer is 12. For example, 13 would not make the given point satisfy the equation. Exam tip: in \(y=mx+c\), identify the \(y\)-intercept directly as \(c\).

Substitute the point coordinates \(x=-4\) and \(y=9\) into the line equation: \(9=-4m-11\). Adding \(11\) to both sides gives \(20=-4m\), so \(m=-5\). Choosing \(5\) would result from a sign error. Exam tip: carefully track the negative sign when substituting a negative value of \(x\).

Writing \(2x+y=5\) in terms of y gives \(y=-2x+5\). Thus, the slope \(-2\) is negative and the y-intercept \(5\) is positive. Exam tip: rewrite the equation as \(y=mx+c\) and check the signs of \(m\) and \(c\).

Writing the equation in terms of y gives \(y=\frac{36}{a}x-\frac{60}{a}\). Hence, the slope is \(\frac{36}{a}\). Since the given slope is 6, \(\frac{36}{a}=6\), so \(a=6\). If \(a=4\), the slope would be 9, so that option is incorrect. Exam tip: In slope-intercept form, the coefficient of x is the slope.

To find the \(y\)-intercept, put \(x=0\). Then \(by+c=0\), so \(y=-\frac{c}{b}\). The expression \(-\frac{a}{b}\) gives the slope, not the intercept. Exam tip: set \(x=0\) first.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope \(m\). Here, the slope is \(4p-3\). Equating it to the given slope, \(4p-3=25\), gives \(4p=28\) and hence \(p=7\). Therefore, option D is correct. If \(p=8\), the slope would be \(29\), not \(25\). Exam tip: compare the coefficient of \(x\) to find the slope.

A line has the form \(y=mx+c\), where \(c\) is the y-intercept. Here, \(c=r-4\). Since the line passes through the origin \((0,0)\), its y-intercept must be zero. Thus, \(r-4=0\), so \(r=4\). Therefore, the slope is \(m=r+8=4+8=12\). The value \(8\) is only part of the coefficient; it is not the slope until \(r\) is found. Exam tip: for a line through the origin, set the constant term equal to zero.

For \(x=5\), every point has x-coordinate 5, so the graph is a vertical line and is perpendicular to the x-axis. On the y-axis, \(x=0\), so this line cannot meet it. Exam tip: \(x=\text{constant}\) always represents a vertical line.

To find the y-intercept, put \(x=0\). Then \(10(0)-5y+25=0\), so \(-5y=-25\) and hence \(y=5\). Therefore, the line meets the y-axis at \((0,5)\). The option \(-5\) is a common sign-error distractor. Exam tip: always set \(x=0\) to find the y-intercept.

In the form \(y=mx+c\), \(m\) is the slope. The slope of the first line is \(q-6\), while that of the second line is \(10\). For equal slopes, \(q-6=10\), so \(q=16\). If \(q=14\), the first slope would be \(8\), not \(10\). Exam tip: In \(y=mx+c\), identify the coefficient of \(x\) as the slope.

In the form y = mx + c, the y-intercept is the constant term c because putting x = 0 gives y = c. The y-intercept of y = -9x + s is s, while that of y = 6x - 17 is -17. Since the intercepts are equal, s = -17. The value -9 is the slope of the first line, not its y-intercept. Exam tip: To find a y-intercept, substitute x = 0.

In the standard form (y=mx+c), (m) is the slope and (c) is the y-intercept. Here, the slope is (a-7) and the y-intercept is (a). Their difference is (a-7)-a=-7, so the terms containing (a) cancel. Hence, the condition is true for every value of (a). Substituting (a=7) merely makes the slope 0; it does not make the condition true only for that value. Exam tip: Compare the equation with (y=mx+c) first, identify (m) and (c), and then calculate the required difference.

Substituting the point \((2,19)\) into \(y=6x+b\) gives \(19=6\times2+b\). Hence, \(b=19-12=7\). As a check, when \(x=5\), \(y=30+7=37\), so the second point also lies on the line. Exam tip: substitute either given point in the line equation to find \(b\).

Writing \(2x+y=0\) as \(y=-2x\) gives slope \(-2\) and y-intercept \(0\). Option B has a positive slope. Exam tip: rewrite a line in \(y=mx+c\) form before identifying its properties.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope \(m\). The first line has slope \(k-8\), while the second has slope \(5\). For equal slopes, \(k-8=5\), so \(k=13\). Hence, option D is correct. The term \(k-2\) is a constant term that gives the y-intercept; it does not affect the slope. Exam tip: to identify the slope, look at the coefficient of \(x\).

In the form \(y=ax+b\), the \(y\)-intercept is the constant term \(b\). The first line has y-intercept \(-8\), while the second has y-intercept \(m-12\). For equal intercepts, \(m-12=-8\), so \(m=4\). The term \(m+2\) belongs to the slope, not the intercept. Exam tip: put \(x=0\) to find a y-intercept.

In \(x=5\), the x-coordinate remains 5 for every point, so the line is vertical. A vertical line has undefined slope and does not meet the y-axis. Exam tip: identify \(x=\text{constant}\) as a vertical line.

The slope is \(\frac{7}{5}\), so an increase of 5 in \(x\) produces an increase of 7 in \(y\). Hence, \(\Delta y=\frac{7}{5}\Delta x\). Putting \(\Delta y=28\), we get \(28=\frac{7}{5}\Delta x\), so \(\Delta x=28\times\frac{5}{7}=20\). Therefore, 20 is correct. If \(x\) increased by 25, \(y\) would increase by 35, not 28. Exam tip: For a linear equation \(y=mx+c\), use \(\Delta y=m\Delta x\) to compare changes.

In the form \(y=px+q\), \(p\) is the slope and \(q\) is the y-intercept. Here, \(p=-8<0\), so as \(x\) increases, \(y\) decreases and the line falls from left to right. Also, \(q=11>0\), so the line meets the y-axis at the positive point \((0,11)\). Therefore, option B is correct. Option C is incorrect because a horizontal line has slope 0, whereas the slope here is \(-8\). Exam tip: In \(y=mx+c\), the sign of \(m\) shows the direction of the line, while \(c\) gives the y-intercept.

In \(y=-4\), there is no \(x\)-term, so the line is horizontal and its slope is 0. It meets the y-axis at \(-4\). In contrast, \(x=-4\) is a vertical line with undefined slope. Exam tip: any equation of the form \(y=c\) represents a horizontal line.