Slope and y-intercept

In the form \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. Equal slopes give \(t+5=14\), so \(t=9\). The y-intercept of the second line is \(2t-7\); hence \(2(9)-7=11\). Option 9 is the value of \(t\), not the y-intercept. Exam tip: first equate the slope coefficients to find the parameter, then substitute it into the constant term.

In the form \(y=mx+c\), \(c\) is the \(y\)-intercept. Since the two lines have the same \(y\)-intercept, \(v-9=8\). Thus, \(v=17\). The slope of the second line \(y=(v+4)x+8\) is \(v+4=17+4=21\). The value 19 would be \(v+2\), not the required slope. Exam tip: in \(y=mx+c\), the coefficient of \(x\) is the slope.

The slope is \(-\frac{7}{4}\). Thus, when \(x\) increases by 4 units, \(y\) decreases by 7 units. For a decrease of 42 in \(y\), \(\frac{7}{4}\Delta x=42\). Hence \(\Delta x=42\times\frac{4}{7}=24\). If 28 were chosen, the decrease in \(y\) would be \(\frac{7}{4}\times 28=49\), not 42. Exam tip: In change-based questions, use the slope; the constant term 18 does not affect the change.

At \(d=0\), \(T=80\), so ₹80 is the y-intercept or fixed initial fee. Increasing \(d\) by 1 km adds ₹12.5, so it is the slope. Treating ₹12.5 as the intercept confuses the coefficient with the constant. Exam tip: set the input to zero to find the intercept.

Given \(c=-15\) and \(m-c=26\). Substituting \(c\) gives \(m-(-15)=26\), so \(m+15=26\). Hence, \(m=11\), and the slope is 11. For example, 10 is not correct because \(10-(-15)=25\), not 26. Exam tip: subtracting a negative number is the same as adding it.

In the line form \(y=mx+c\), \(c\) is the value of the \(y\)-intercept. Substituting \(m=-5\) into \(3c+m=31\) gives \(3c-5=31\). Hence, \(3c=36\), so \(c=12\). Therefore, the \(y\)-intercept is 12. Choosing 11 would be incorrect because it does not satisfy \(3c-5=31\). Exam tip: substitute the given value of \(m\) into the relation first, then solve for \(c\).

\(x=3\) is a vertical line. Here \(\Delta x=0\), so \(m=\frac{\Delta y}{\Delta x}\) is undefined. It does not meet the \(y\)-axis, where \(x=0\). In contrast, \(y=3\) has zero slope. Exam tip: a line of the form \(x=\text{constant}\) is vertical.

Dividing both sides of the equation by 8 gives y=3x-8. For the y-intercept, put x=0, which gives y=-8. Therefore, the y-intercept is -8. Although -64 is the constant term in the original equation, it must be divided by 8 before identifying the y-intercept. Exam tip: In the form y=mx+c, c is the y-intercept.

In \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept. For option A, \(m=3\) is positive and \(c=-5\) is negative. Option B has a negative intercept but its slope is negative. Exam tip: check the signs of \(m\) and \(c\) separately.

Substitute the point \((-5,12)\) into \(y=mx-13\): \(12=m(-5)-13\). Hence, \(25=-5m\), so \(m=-5\). Therefore, the correct option is \(\mathbf{-5}\). If \(m=5\), the equation gives \(y=-38\) when \(x=-5\), so it cannot pass through the given point. Exam tip: To find an unknown slope parameter, substitute the coordinates of the given point into the line equation.

Rewriting \(2x-y=4\) as \(y=2x-4\) gives slope \(m=2\), which is positive, and y-intercept \(c=-4\), which is negative. Although \(-2x+y=4\) also has a positive slope, its intercept is \(4\). Exam tip: convert to \(y=mx+c\) before identifying signs.

The given line has slope \(3/2\). Option A gives \(y=(3/2)x-4\), so it is parallel and has intercept \(-4\). C has intercept \(-4\) but slope \(2/3\). Tip: compare \(m\) in \(y=mx+c\).

\(x=4\) is a vertical line, so its slope is undefined. It cannot meet the \(y\)-axis because every point on the \(y\)-axis has \(x=0\). In contrast, \(y=-3\) has slope 0. Exam tip: \(x=\text{constant}\) represents a vertical line.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope. Here, the slope is \(5p-2\), so \(5p-2=38\). Thus, \(5p=40\) and \(p=8\). If \(p=7\), the slope would be \(33\), so it is not correct. Exam tip: identify the coefficient of \(x\) to find the slope.

Since the line passes through the origin \((0,0)\), put \(x=0\) and \(y=0\). Its constant term must be zero: \(r-6=0\). Hence, \(r=6\). The slope is the coefficient of \(x\), so \(r+10=6+10=16\). Option \(6\) is the value of \(r\), not the slope. Exam tip: in \(y=mx+c\), \(m\) is the slope and \(c\) is the y-intercept.

Comparing with \(y=mx+c\), we get \(m=-3\) and \(c=4\). A negative slope means the line falls as we move right, and its \(y\)-intercept is 4. Exam tip: \(c\) always gives the \(y\)-axis intercept.

To find the y-intercept, put \(x=0\). This gives \(-6y+30=0\), so \(-6y=-30\) and \(y=5\). Hence, the y-intercept is \(5\). The value \(-5\) can result from a sign error. Exam tip: always set \(x=0\) to find the y-intercept.

In the form \(y=mx+c\), the coefficient of \(x\) is the slope \(m\). The slopes of the first and second lines are \(q-9\) and \(12\), respectively. For equal slopes, \(q-9=12\), so \(q=21\). If 20 were chosen, the first slope would be 11, not 12. Exam tip: for parallel lines or lines with equal slopes, compare the coefficients of \(x\).

In the form \(y=mx+c\), the constant term \(c\) is the \(y\)-intercept. The \(y\)-intercept of \(y=-10x+s\) is \(s\), while that of \(y=7x-19\) is \(-19\). Since the intercepts are equal, \(s=-19\). Note that \(-10\) is the slope of the first line, not its \(y\)-intercept. Exam tip: Put \(x=0\) to find the \(y\)-intercept.

For a line, the coefficient of x is its slope, so the slope is a+8. The constant term is the y-intercept, so the y-intercept is a. Their difference is (a+8)-a=8, which is true for every real value of a. Values such as a=8 or a=0 are only particular cases; the condition is not restricted to them. Exam tip: In y=mx+c, m is the slope and c is the y-intercept.