This question is from "Mathematics for Economists" by Simon and Blume.
IS curve: $[1-c_1(1-t_1)-a_0]Y+(a+c_2)r=c_0-c_1t+I^*+G$
LM curve: $mY-hr=M_s-M^*$
The parameters $c_1$,$t_1$ and $a_0$ are between $0$ and $1$. It is usually assumed that $0<c_1(1-t_1)+a_0<1$, so that the normal vector points northeast and the IS-line has negative slope $$-\frac{a+c_2}{1-c_1(1-t_1)-a_0}.$$ The normal vector to LM-line is $(m,-h)$ which point southeast because $h,m>0$ and so LM-linehas a positive slope $\frac{h}{m}$. Now,if $G$ or $I^*$ increases or if $t_0$ decreases, then the right hand side of the IS- equation increases and IS-line shifts outward as in Figure 10.34.The result is an increase in the equilibrium $Y$ and $r$.
Note that this result would hold even if slope of the IS-line were positive, as long as it was less than the slope of the LM-line.
But, when I draw pictures, I find that the opposite is true, that is, when IS-line has its slope less than LM, increasing $I^*$ or $G$ shifts IS-line downward and thus causes decrease in equilibrium $Y$ and $r$.Also,when IS-line has its slope greater than LM, increasing $I^*$ or $G$ shifts IS-line outward and thus causes decrease in equilibrium $Y$ and $r$.
Where am I wrong?(Or is it a typo?)
An increase in G or I* shifts the IS curve up. As long as the slope of the LM curve is greater than the IS curve, the change in Y and r is positive.
EDIT:
To see this, rewrite the IS-LM equations in terms of $r$, \begin{align*} r_{is} &= -\frac{(1-c_1(1 - t_1) - a_0)}{a + c_2} Y + \frac{(c_0 - c_1t_0 + I^* + G)}{a + c_2} \\ r_{lm} &= \frac{m}{h}Y - \frac{(M_s - M^*)}{h}. \end{align*}
So, any increase in $G$ or $I^*$ will result in an increase in $r_{is}$ for all values of $Y$.