\displaystyle \textbf{Note: }\text{The equation of a line in intercept form is }\frac{x}{a}+\frac{y}{b}=1,
\displaystyle \text{where }a\text{ is the }x\text{-intercept and }b\text{ is the }y\text{-intercept.}

\displaystyle \textbf{Question 1: }\text{Find the equation of the straight line:}
\displaystyle \text{i) cutting off intercepts }3\text{ and }2\text{ on the axes.}
\displaystyle \text{ii) cutting off intercepts }-5\text{ and }6\text{ on the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{i) Here }a=3\text{ and }b=2.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{3}+\frac{y}{2}=1.
\displaystyle 2x+3y=6.
\displaystyle \therefore 2x+3y-6=0.
\displaystyle \text{ii) Here }a=-5\text{ and }b=6.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{-5}+\frac{y}{6}=1.
\displaystyle -6x+5y=30.
\displaystyle \therefore 6x-5y+30=0.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the equation of the straight line which passes through }(1,-2)
\displaystyle \text{and cuts off equal intercepts on the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Since the intercepts are equal, let }a=b.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{a}+\frac{y}{a}=1.
\displaystyle \therefore x+y=a.
\displaystyle \text{Since the line passes through }(1,-2),
\displaystyle 1+(-2)=a.
\displaystyle \therefore a=-1.
\displaystyle \therefore \text{The required equation is }x+y=-1,
\displaystyle \text{or }x+y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the equation of the straight line which passes through }(5,6)\text{ and has}
\displaystyle \text{intercepts on the axes: (i) equal in magnitude and both positive,}
\displaystyle \text{(ii) equal in magnitude but opposite in sign.}
\displaystyle \text{Answer:}
\displaystyle \text{i) Since the intercepts are equal and positive, let }a=b.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{a}+\frac{y}{a}=1.
\displaystyle \therefore x+y=a.
\displaystyle \text{Since the line passes through }(5,6),
\displaystyle 5+6=a.
\displaystyle \therefore a=11.
\displaystyle \therefore \text{The required equation is }x+y=11.
\displaystyle \text{ii) Since the intercepts are equal in magnitude but opposite in sign, let }b=-a.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{a}+\frac{y}{-a}=1.
\displaystyle \therefore x-y=a.
\displaystyle \text{Since the line passes through }(5,6),
\displaystyle 5-6=a.
\displaystyle \therefore a=-1.
\displaystyle \therefore \text{The required equation is }x-y=-1,
\displaystyle \text{or }x-y+1=0.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{For what values of }a\text{ and }b\text{ are the intercepts cut off on the coordinate}
\displaystyle \text{axes by the line }ax+by+8=0\text{ equal in length but opposite in sign to those}
\displaystyle \text{cut off by the line }2x-3y+6=0\text{ on the axes?}
\displaystyle \text{Answer:}
\displaystyle \text{Given line }2x-3y+6=0.
\displaystyle 2x-3y=-6.
\displaystyle \frac{x}{-3}+\frac{y}{2}=1.
\displaystyle \therefore \text{Its }x\text{-intercept is }-3\text{ and its }y\text{-intercept is }2.
\displaystyle \text{The other line is }ax+by+8=0.
\displaystyle ax+by=-8.
\displaystyle \frac{x}{-\frac{8}{a}}+\frac{y}{-\frac{8}{b}}=1.
\displaystyle \therefore \text{Its }x\text{-intercept is }-\frac{8}{a}\text{ and its }y\text{-intercept is }-\frac{8}{b}.
\displaystyle \text{Since the corresponding intercepts are equal in length but opposite in sign,}
\displaystyle -\frac{8}{a}=-(-3)=3.
\displaystyle \therefore a=-\frac{8}{3}.
\displaystyle \text{Also,}
\displaystyle -\frac{8}{b}=-(2)=-2.
\displaystyle \therefore b=4.
\displaystyle \therefore a=-\frac{8}{3}\text{ and }b=4.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the equation of the straight line which cuts off equal positive intercepts}
\displaystyle \text{on the coordinate axes and whose product is }25.
\displaystyle \text{Answer:}
\displaystyle \text{Let the equal positive intercepts be }a\text{ and }a.
\displaystyle \text{Their product is }25.
\displaystyle a^2=25.
\displaystyle a=\pm5.
\displaystyle \text{Since the intercepts are positive, }a=5.
\displaystyle \text{Using the intercept form,}
\displaystyle \frac{x}{5}+\frac{y}{5}=1.
\displaystyle \therefore x+y=5.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the equation of the line which passes through the point }(-4,3)\text{ and such}
\displaystyle \text{that this point divides internally, in the ratio }5:3,\text{ the portion of the line}
\displaystyle \text{intercepted between the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{The point }P(-4,3)\text{ divides }AB\text{ internally in the ratio }5:3.
\displaystyle \text{By the section formula,}
\displaystyle -4=\frac{5(0)+3a}{5+3}.
\displaystyle -4=\frac{3a}{8}.
\displaystyle \therefore a=-\frac{32}{3}.
\displaystyle \text{Similarly,}
\displaystyle 3=\frac{5b+3(0)}{5+3}.
\displaystyle 3=\frac{5b}{8}.
\displaystyle \therefore b=\frac{24}{5}.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{-\frac{32}{3}}+\frac{y}{\frac{24}{5}}=1.
\displaystyle -\frac{3x}{32}+\frac{5y}{24}=1.
\displaystyle -9x+20y=96.
\displaystyle \therefore 9x-20y+96=0.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{A straight line passes through the point }(\alpha,\beta),\text{ and this point bisects}
\displaystyle \text{the portion of the line intercepted between the coordinate axes. Show that the equation}
\displaystyle \text{of the straight line is }\frac{x}{2\alpha}+\frac{y}{2\beta}=1.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Since }(\alpha,\beta)\text{ bisects }AB,\text{ it is the midpoint of }AB.
\displaystyle \therefore \alpha=\frac{a+0}{2}.
\displaystyle \therefore a=2\alpha.
\displaystyle \text{Similarly,}
\displaystyle \beta=\frac{0+b}{2}.
\displaystyle \therefore b=2\beta.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \therefore \frac{x}{2\alpha}+\frac{y}{2\beta}=1.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the equation of the line which passes through the point }(3,4)\text{ and is such}
\displaystyle \text{that the portion intercepted between the coordinate axes is divided by the point in the}
\displaystyle \text{ratio }2:3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given that }P(3,4)\text{ divides }AB\text{ internally in the ratio }2:3.
\displaystyle \text{Thus, }AP:PB=2:3.
\displaystyle \text{By the section formula,}
\displaystyle 3=\frac{2(0)+3a}{2+3}.
\displaystyle 3=\frac{3a}{5}.
\displaystyle \therefore a=5.
\displaystyle \text{Similarly,}
\displaystyle 4=\frac{2b+3(0)}{2+3}.
\displaystyle 4=\frac{2b}{5}.
\displaystyle \therefore b=10.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{5}+\frac{y}{10}=1.
\displaystyle \therefore 2x+y=10.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The point }R(h,k)\text{ divides a line segment intercepted between the coordinate}
\displaystyle \text{axes internally in the ratio }1:2.\text{ Find the equation of the line.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given that }R(h,k)\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle \text{Thus, }AR:RB=1:2.
\displaystyle \text{By the section formula,}
\displaystyle h=\frac{1(0)+2a}{1+2}=\frac{2a}{3}.
\displaystyle \therefore a=\frac{3h}{2}.
\displaystyle \text{Similarly,}
\displaystyle k=\frac{1(b)+2(0)}{1+2}=\frac{b}{3}.
\displaystyle \therefore b=3k.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{\frac{3h}{2}}+\frac{y}{3k}=1.
\displaystyle \frac{2x}{3h}+\frac{y}{3k}=1.
\displaystyle \therefore 2kx+hy=3hk.
\displaystyle \therefore 2kx+hy-3hk=0.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the equation of the straight line which passes through the point }(-3,8)
\displaystyle \text{and cuts off positive intercepts on the coordinate axes whose sum is }7.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Since the intercepts are positive and their sum is }7,
\displaystyle a+b=7.
\displaystyle \therefore b=7-a.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Since the line passes through }(-3,8),
\displaystyle \frac{-3}{a}+\frac{8}{7-a}=1.
\displaystyle -3(7-a)+8a=a(7-a).
\displaystyle -21+3a+8a=7a-a^2.
\displaystyle a^2+4a-21=0.
\displaystyle (a-3)(a+7)=0.
\displaystyle \therefore a=3\text{ or }a=-7.
\displaystyle \text{Since the }x\text{-intercept is positive, }a=3.
\displaystyle \therefore b=7-3=4.
\displaystyle \text{Hence, the equation of the line is}
\displaystyle \frac{x}{3}+\frac{y}{4}=1.
\displaystyle \therefore 4x+3y=12.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the equation of the straight line which passes through the point }(-4,3)
\displaystyle \text{and is such that the portion intercepted between the coordinate axes is divided by the}
\displaystyle \text{point internally in the ratio }5:3.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{The point }P(-4,3)\text{ divides }AB\text{ internally in the ratio }5:3.
\displaystyle \text{Thus, }AP:PB=5:3.
\displaystyle \text{By the section formula,}
\displaystyle -4=\frac{5(0)+3a}{5+3}.
\displaystyle -4=\frac{3a}{8}.
\displaystyle \therefore a=-\frac{32}{3}.
\displaystyle \text{Similarly,}
\displaystyle 3=\frac{5b+3(0)}{5+3}.
\displaystyle 3=\frac{5b}{8}.
\displaystyle \therefore b=\frac{24}{5}.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{-\frac{32}{3}}+\frac{y}{\frac{24}{5}}=1.
\displaystyle -\frac{3x}{32}+\frac{5y}{24}=1.
\displaystyle -9x+20y=96.
\displaystyle \therefore 9x-20y+96=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find the equations of the lines which pass through the point }(22,-6)\text{ and}
\displaystyle \text{are such that the }x\text{-intercept exceeds the }y\text{-intercept by }5.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Since the }x\text{-intercept exceeds the }y\text{-intercept by }5,
\displaystyle a=b+5.
\displaystyle \therefore b=a-5.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Since the line passes through }(22,-6),
\displaystyle \frac{22}{a}-\frac{6}{a-5}=1.
\displaystyle 22(a-5)-6a=a(a-5).
\displaystyle 22a-110-6a=a^2-5a.
\displaystyle a^2-21a+110=0.
\displaystyle (a-11)(a-10)=0.
\displaystyle \therefore a=11\text{ or }a=10.
\displaystyle \text{When }a=11,\quad b=11-5=6.
\displaystyle \text{When }a=10,\quad b=10-5=5.
\displaystyle \text{Therefore, the equations of the lines are}
\displaystyle \frac{x}{11}+\frac{y}{6}=1\quad\text{or}\quad\frac{x}{10}+\frac{y}{5}=1.
\displaystyle \therefore 6x+11y=66\quad\text{or}\quad x+2y=10.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the equation of the line which passes through }P(1,-7)\text{ and meets the}
\displaystyle \text{coordinate axes at }A\text{ and }B\text{ respectively such that }4AP-3BP=0.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given }4AP-3BP=0.
\displaystyle \therefore 4AP=3BP.
\displaystyle \therefore AP:PB=3:4.
\displaystyle \text{Thus, }P(1,-7)\text{ divides }AB\text{ internally in the ratio }3:4.
\displaystyle \text{By the section formula,}
\displaystyle 1=\frac{3(0)+4a}{3+4}=\frac{4a}{7}.
\displaystyle \therefore a=\frac{7}{4}.
\displaystyle \text{Similarly,}
\displaystyle -7=\frac{3b+4(0)}{3+4}=\frac{3b}{7}.
\displaystyle \therefore b=-\frac{49}{3}.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{\frac{7}{4}}+\frac{y}{-\frac{49}{3}}=1.
\displaystyle \frac{4x}{7}-\frac{3y}{49}=1.
\displaystyle \therefore 28x-3y=49.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Find the equations of the lines passing through the point }(2,2)\text{ and}
\displaystyle \text{cutting off intercepts on the coordinate axes whose sum is }9.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given that the sum of the intercepts is }9,
\displaystyle a+b=9.
\displaystyle \therefore b=9-a.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Since the line passes through }(2,2),
\displaystyle \frac{2}{a}+\frac{2}{9-a}=1.
\displaystyle 2(9-a)+2a=a(9-a).
\displaystyle 18=9a-a^2.
\displaystyle a^2-9a+18=0.
\displaystyle (a-3)(a-6)=0.
\displaystyle \therefore a=3\text{ or }a=6.
\displaystyle \text{When }a=3,\quad b=9-3=6.
\displaystyle \text{When }a=6,\quad b=9-6=3.
\displaystyle \text{Therefore, the equations of the lines are}
\displaystyle \frac{x}{3}+\frac{y}{6}=1\quad\text{or}\quad\frac{x}{6}+\frac{y}{3}=1.
\displaystyle \therefore 2x+y=6\quad\text{or}\quad x+2y=6.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Find the equation of the straight line which passes through }P(2,6)\text{ and cuts the}
\displaystyle \text{coordinate axes at }A\text{ and }B\text{ respectively such that }\frac{AP}{BP}=\frac{2}{3}.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given }AP:PB=2:3.
\displaystyle \text{Thus, }P(2,6)\text{ divides }AB\text{ internally in the ratio }2:3.
\displaystyle \text{By the section formula,}
\displaystyle 2=\frac{2(0)+3a}{2+3}=\frac{3a}{5}.
\displaystyle \therefore a=\frac{10}{3}.
\displaystyle \text{Similarly,}
\displaystyle 6=\frac{2b+3(0)}{2+3}=\frac{2b}{5}.
\displaystyle \therefore b=15.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{\frac{10}{3}}+\frac{y}{15}=1.
\displaystyle \frac{3x}{10}+\frac{y}{15}=1.
\displaystyle 9x+2y=30.
\displaystyle \therefore 9x+2y-30=0.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Find the equations of the straight lines, each of which passes through the point}
\displaystyle (3,2)\text{ and cuts off intercepts }a\text{ and }b\text{ on the }x\text{-axis and }y\text{-axis respectively,}
\displaystyle \text{such that }a-b=2.
\displaystyle \text{Answer:}
\displaystyle \text{Let the line meet the }x\text{-axis at }A(a,0)\text{ and the }y\text{-axis at }B(0,b).
\displaystyle \text{Given, }a-b=2.
\displaystyle \therefore a=b+2.
\displaystyle \text{Using the intercept form, the equation of the line is}
\displaystyle \frac{x}{a}+\frac{y}{b}=1.
\displaystyle \text{Since the line passes through }(3,2),
\displaystyle \frac{3}{b+2}+\frac{2}{b}=1.
\displaystyle 3b+2(b+2)=b(b+2).
\displaystyle 3b+2b+4=b^2+2b.
\displaystyle b^2-3b-4=0.
\displaystyle (b+1)(b-4)=0.
\displaystyle \therefore b=-1\text{ or }b=4.
\displaystyle \text{When }b=-1,\quad a=-1+2=1.
\displaystyle \text{When }b=4,\quad a=4+2=6.
\displaystyle \text{Therefore, the equations of the lines are}
\displaystyle \frac{x}{1}+\frac{y}{-1}=1\quad\text{or}\quad\frac{x}{6}+\frac{y}{4}=1.
\displaystyle \therefore x-y=1\quad\text{or}\quad 2x+3y=12.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the equations of the straight lines which pass through the origin and trisect}
\displaystyle \text{the portion of the straight line }2x+3y=6\text{ intercepted between the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }2x+3y=6.
\displaystyle \text{Its }x\text{-intercept is }A(3,0)\text{ and its }y\text{-intercept is }B(0,2).
\displaystyle \text{Let }P\text{ and }Q\text{ be the trisection points of the line segment }AB.
\displaystyle \text{The point }P\text{ divides }AB\text{ internally in the ratio }1:2.
\displaystyle P=\left(\frac{1(0)+2(3)}{1+2},\frac{1(2)+2(0)}{1+2}\right)2021-01-11_10-45-56
\displaystyle =\left(2,\frac{2}{3}\right).
\displaystyle \text{The point }Q\text{ divides }AB\text{ internally in the ratio }2:1.
\displaystyle Q=\left(\frac{2(0)+1(3)}{2+1},\frac{2(2)+1(0)}{2+1}\right)
\displaystyle =\left(1,\frac{4}{3}\right).
\displaystyle \text{Slope of }OP=\frac{\frac{2}{3}-0}{2-0}=\frac{1}{3}.
\displaystyle \text{Therefore, the equation of }OP\text{ is}
\displaystyle y-0=\frac{1}{3}(x-0).
\displaystyle \therefore x-3y=0.
\displaystyle \text{Slope of }OQ=\frac{\frac{4}{3}-0}{1-0}=\frac{4}{3}.
\displaystyle \text{Therefore, the equation of }OQ\text{ is}
\displaystyle y-0=\frac{4}{3}(x-0).
\displaystyle \therefore 4x-3y=0.
\displaystyle \text{Hence, the required equations are }x-3y=0\text{ and }4x-3y=0.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find the equation of the straight line passing through the point }(2,1)\text{ and}
\displaystyle \text{bisecting the portion of the straight line }3x-5y=15\text{ lying between the axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }3x-5y=15.
\displaystyle \text{Putting }y=0,\text{ we get }x=5.
\displaystyle \therefore \text{The }x\text{-intercept is }A(5,0).
\displaystyle \text{Putting }x=0,\text{ we get }y=-3.
\displaystyle \therefore \text{The }y\text{-intercept is }B(0,-3).
\displaystyle \text{Let }M\text{ be the midpoint of }AB.
\displaystyle M=\left(\frac{5+0}{2},\frac{0+(-3)}{2}\right)=\left(\frac{5}{2},-\frac{3}{2}\right).
\displaystyle \text{The required line passes through }P(2,1)\text{ and }M\left(\frac{5}{2},-\frac{3}{2}\right).
\displaystyle \text{Its slope is}
\displaystyle m=\frac{1-\left(-\frac{3}{2}\right)}{2-\frac{5}{2}}=\frac{\frac{5}{2}}{-\frac{1}{2}}=-5.
\displaystyle \text{Therefore, the equation of the required line is}
\displaystyle y-1=-5(x-2).
\displaystyle y-1=-5x+10.
\displaystyle \therefore 5x+y-11=0.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Find the equation of the straight line passing through the origin and bisecting}
\displaystyle \text{the portion of the line }ax+by+c=0\text{ intercepted between the coordinate axes.}
\displaystyle \text{Answer:}
\displaystyle \text{The given line is }ax+by+c=0.
\displaystyle \text{Putting }y=0,\text{ we get }x=-\frac{c}{a}.
\displaystyle \therefore \text{The }x\text{-intercept is }A\left(-\frac{c}{a},0\right).
\displaystyle \text{Putting }x=0,\text{ we get }y=-\frac{c}{b}.
\displaystyle \therefore \text{The }y\text{-intercept is }B\left(0,-\frac{c}{b}\right).
\displaystyle \text{Let }M\text{ be the midpoint of }AB.
\displaystyle M=\left(\frac{-\frac{c}{a}+0}{2},\frac{0-\frac{c}{b}}{2}\right)
\displaystyle =\left(-\frac{c}{2a},-\frac{c}{2b}\right).
\displaystyle \text{The required line passes through the origin }O(0,0)\text{ and the midpoint }M.
\displaystyle \text{Slope of }OM=\frac{-\frac{c}{2b}-0}{-\frac{c}{2a}-0}=\frac{a}{b}.
\displaystyle \text{Therefore, the equation of }OM\text{ is}
\displaystyle y-0=\frac{a}{b}(x-0).
\displaystyle by=ax.
\displaystyle \therefore ax-by=0.
\displaystyle \\

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